Definite Integration
238 JEE Maths previous year questions on Definite Integration — options free on every question; 24 include the answer & explanation free, the rest unlock with PYQ Pass.
The value of the definite integral \({\int }_{0}^{2}\frac{1}{{3}^{x}+3}dx\) is
[JEE Advanced 2026]
\(\frac{1}{3}\)
Given
\(\displaystyle I=\int_0^2 \frac{1}{3^x+3}\,dx\)
Let \(3^x=t\)
\(\Rightarrow dx=\dfrac{dt}{t\ln3}\)
When \(x=0,\; t=1\)
When \(x=2,\; t=9\)
Therefore,
\(\displaystyle I=\frac1{\ln3}\int_1^9 \frac{1}{t(t+3)}\,dt\)
\(=\frac1{\ln3}\int_1^9\left(\frac1{3t}-\frac1{3(t+3)}\right)dt\)
\(=\frac1{3\ln3}\left[\ln t-\ln(t+3)\right]_1^9\)
\(=\frac1{3\ln3}\left[\ln\frac{9}{12}-\ln\frac14\right]\)
\(=\frac1{3\ln3}\ln\left(\frac34\cdot4\right)\)
\(=\frac1{3\ln3}\ln3\)
\(=\frac13\)
Hence,
\(\displaystyle \int_0^2 \frac{1}{3^x+3}\,dx=\frac13\)
\(4{\int }_{0}^{1}\left(\frac{1}{\sqrt{3+{\mathrm{x}}^{2}}+\sqrt{1+{\mathrm{x}}^{2}}}\right)\mathrm{dx}-3{\log }_{\mathrm{e}}\left(\sqrt{3}\right)\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
\(2-\sqrt{2}-{\log }_{\mathrm{e}}(1+\sqrt{2})\)
Let \(I=4\int_0^1\left(\frac{1}{\sqrt{3+x^2}+\sqrt{1+x^2}}\right)dx-3\log_e(\sqrt{3})\).
\(I=2\int_0^1\left(\sqrt{x^2+3}-\sqrt{x^2+1}\right)dx-3\log_e(\sqrt{3})\).
Now \( \int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log_e\left(x+\sqrt{x^2+a^2}\right) \).
Therefore \( \int_0^1 \sqrt{x^2+3}dx=\left[\frac{x}{2}\sqrt{x^2+3}+\frac{3}{2}\log_e\left(x+\sqrt{x^2+3}\right)\right]_0^1 \).
So \( \int_0^1 \sqrt{x^2+3}dx=1+\frac{3}{2}\log_e\left(\frac{3}{\sqrt{3}}\right)=1+\frac{3}{2}\log_e(\sqrt{3}) \).
Also \( \int_0^1 \sqrt{x^2+1}dx=\left[\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log_e\left(x+\sqrt{x^2+1}\right)\right]_0^1 \).
So \( \int_0^1 \sqrt{x^2+1}dx=\frac{\sqrt{2}}{2}+\frac{1}{2}\log_e(1+\sqrt{2}) \).
Hence \(I=2\left(1+\frac{3}{2}\log_e(\sqrt{3})-\frac{\sqrt{2}}{2}-\frac{1}{2}\log_e(1+\sqrt{2})\right)-3\log_e(\sqrt{3})\).
So \(I=2-\sqrt{2}+3\log_e(\sqrt{3})-\log_e(1+\sqrt{2})-3\log_e(\sqrt{3})\).
Therefore \(I=2-\sqrt{2}-\log_e(1+\sqrt{2})\).
Let \(f:R\to R\) be a twice differentiable function such that \(f(2)=1\). If \(F(x)=xf(x)\) for all \(x\in R,\text{ }{\int }_{0}^{2}x{F}^{'}\left(x\right)dx=6\) and \({\int }_{0}^{2}{x}^{2}{\mathrm{F}}^{"}\left(x\right)\mathrm{d}x=40\), then \({F}^{'}\left(2\right)+{\int }_{0}^{2}F\left(x\right)dx\) is equal to :
[JEE Main 2025, 28 Jan (Shift 2)]
11
\(F\left(x\right)=xf\left(x\right)\\ \Rightarrow F\left(x\right)=2f\left(2\right)=2\)
\(\text{Given that}{\int }_{0}^{2}x{F}^{'}\left(x\right)dx=6\\ ={\left.xF\left(x\right)\right|}_{0}^{2}-{\int }_{0}^{2}F\left(x\right)dx=6\left[\text{using integration by parts}\right]\\ =2\mathrm{F}(2)-{\int }_{0}^{2}xf\left(x\right)\mathrm{dx}=6\\ \Rightarrow {\int }_{0}^{2}\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=-2....\left(1\right)\\ \mathrm{Also}\\ {\int }_{0}^{2}{x}^{2}{F}^{"}\left(x\right)dx={\left.{x}^{2}{F}^{'}\left(x\right)\right|}_{0}^{2}-2{\int }_{0}^{2}x{F}^{'}\left(x\right)dx=40\\ =4{\mathrm{F}}^{'}\left(2\right)-2\times 6=40\\ {F}^{'}\left(2\right)=13\\ ∴{\mathrm{F}}^{'}\left(2\right)+{\int }_{0}^{2}\mathrm{F}\left(\mathrm{x}\right)=13-2\\ =11\)
\({\int }_{0}^{\pi /4}\frac{{\cos }^{2}x{\sin }^{2}x}{{\left({\cos }^{3}x+{\sin }^{3}x\right)}^{2}}dx\) is equal to
[JEE Main 2024, 6 Apr (Shift 1)]
\(\frac{1}{6}\)
Given integral is \(I=\int_0^{\pi/4}\frac{\cos^2x\sin^2x}{(\cos^3x+\sin^3x)^2}\,dx\)
Put \(\tan x=t\)
Then \(\sec^2x\,dx=dt\), so \(dx=\frac{dt}{1+t^2}\)
Also, \(\sin x=\frac{t}{\sqrt{1+t^2}}\) and \(\cos x=\frac{1}{\sqrt{1+t^2}}\)
So, \(\cos^2x\sin^2x=\frac{t^2}{(1+t^2)^2}\)
Now, \(\cos^3x+\sin^3x=\frac{1}{(1+t^2)^{3/2}}+\frac{t^3}{(1+t^2)^{3/2}}\)
Therefore, \(\cos^3x+\sin^3x=\frac{1+t^3}{(1+t^2)^{3/2}}\)
So, \((\cos^3x+\sin^3x)^2=\frac{(1+t^3)^2}{(1+t^2)^3}\)
When \(x=0\), \(t=0\), and when \(x=\frac{\pi}{4}\), \(t=1\)
Therefore, \(I=\int_0^1 \frac{\frac{t^2}{(1+t^2)^2}}{\frac{(1+t^3)^2}{(1+t^2)^3}}\cdot\frac{dt}{1+t^2}\)
\(I=\int_0^1 \frac{t^2}{(1+t^3)^2}\,dt\)
Put \(1+t^3=u\)
Then \(3t^2\,dt=du\), so \(t^2\,dt=\frac{du}{3}\)
When \(t=0\), \(u=1\), and when \(t=1\), \(u=2\)
Therefore, \(I=\frac{1}{3}\int_1^2 \frac{du}{u^2}\)
\(I=\frac{1}{3}\left[-\frac{1}{u}\right]_1^2\)
\(I=\frac{1}{3}\left(-\frac{1}{2}+1\right)\)
\(I=\frac{1}{3}\cdot\frac{1}{2}\)
\(I=\frac{1}{6}\)
Hence, the value of the integral is \(\frac{1}{6}\)
If \(\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}\), where \(a, b, c\) are rational numbers, then \(2 a+3 b-4 c\) is equal to:
[JEE Main 2024, 27 Jan (Shift 1)]
8
Let \(I=\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx\).
\(I=\frac{1}{2}\int_0^1(\sqrt{3+x}-\sqrt{1+x})dx\).
Now \(I=\frac{1}{2}\left[\int_0^1\sqrt{3+x}\,dx-\int_0^1\sqrt{1+x}\,dx\right]\).
Using \(\int \sqrt{x+a}\,dx=\frac{2}{3}(x+a)^{3/2}\),
\(\int_0^1\sqrt{3+x}\,dx=\left[\frac{2}{3}(x+3)^{3/2}\right]_0^1=\frac{2}{3}(8-3\sqrt3)\).
Also \(\int_0^1\sqrt{1+x}\,dx=\left[\frac{2}{3}(x+1)^{3/2}\right]_0^1=\frac{2}{3}(2\sqrt2-1)\).
Thus \(I=\frac{1}{2}\left[\frac{2}{3}(8-3\sqrt3)-\frac{2}{3}(2\sqrt2-1)\right]\).
So \(I=\frac{1}{3}(8-3\sqrt3-2\sqrt2+1)\).
Therefore \(I=3-\frac{2}{3}\sqrt2-\sqrt3\).
Comparing with \(a+b\sqrt2+c\sqrt3\), we get \(a=3\), \(b=-\frac{2}{3}\), and \(c=-1\).
Now \(2a+3b-4c=2(3)+3\left(-\frac{2}{3}\right)-4(-1)\).
So \(2a+3b-4c=6-2+4=8\).
If \(\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=a+b \sqrt{2}+c \sqrt{3}\), where \(a, b, c\) are rational numbers, then \(2 a+3 b-4 c\) is equal to:
[JEE Main 2024, 27 Jan (Shift 1)]
8
Let \(I=\int_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}}dx\).
\(I=\frac{1}{2}\int_0^1(\sqrt{3+x}-\sqrt{1+x})dx\).
Now \(I=\frac{1}{2}\left[\int_0^1\sqrt{3+x}\,dx-\int_0^1\sqrt{1+x}\,dx\right]\).
Using \(\int \sqrt{x+a}\,dx=\frac{2}{3}(x+a)^{3/2}\),
\(\int_0^1\sqrt{3+x}\,dx=\left[\frac{2}{3}(x+3)^{3/2}\right]_0^1=\frac{2}{3}(8-3\sqrt3)\).
Also \(\int_0^1\sqrt{1+x}\,dx=\left[\frac{2}{3}(x+1)^{3/2}\right]_0^1=\frac{2}{3}(2\sqrt2-1)\).
Thus \(I=\frac{1}{2}\left[\frac{2}{3}(8-3\sqrt3)-\frac{2}{3}(2\sqrt2-1)\right]\).
So \(I=\frac{1}{3}(8-3\sqrt3-2\sqrt2+1)\).
Therefore \(I=3-\frac{2}{3}\sqrt2-\sqrt3\).
Comparing with \(a+b\sqrt2+c\sqrt3\), we get \(a=3\), \(b=-\frac{2}{3}\), and \(c=-1\).
Now \(2a+3b-4c=2(3)+3\left(-\frac{2}{3}\right)-4(-1)\).
So \(2a+3b-4c=6-2+4=8\).
The value of \(k \in N\) for which the integral \(I_n=\int_0^1\left(1-x^k\right)^n d x, n \in N\), satisfies \(147 I_{20}=148 I_{21}\) is
[JEE Main 2024, 8 Apr (Shift 1)]
7
\({I}_{n}={\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n}\cdot 1dx\)
\({I}_{n}={\left[{\left(1-{x}^{k}\right)}^{n}\cdot x\right]}_{0}^{1}-nk{\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n-1}\left(-{x}^{k-1}\right)\cdot xdx\)
\(=-kn{\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n-1}\left(1-{x}^{k}-1\right)dx\)
\({I}_{n}=-nk{\int }_{0}^{1}\left[{\left(1-{x}^{k}\right)}^{n}-{\left(1-{x}^{k}\right)}^{n-1}\right]dx\)
\({I}_{n}=nk{I}_{n-1}-nk{I}_{n}\)
\(\frac{{I}_{n}}{{I}_{n-1}}=\frac{nk}{nk+1}\)
\(\frac{{I}_{21}}{{I}_{20}}=\frac{21k}{1+21k}\)
\(=\frac{147}{148}\Rightarrow k=7\)
The value of \(k \in N\) for which the integral \(I_n=\int_0^1\left(1-x^k\right)^n d x, n \in N\), satisfies \(147 I_{20}=148 I_{21}\) is
[JEE Main 2024, 8 Apr (Shift 1)]
7
\({I}_{n}={\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n}\cdot 1dx\)
\({I}_{n}={\left[{\left(1-{x}^{k}\right)}^{n}\cdot x\right]}_{0}^{1}-nk{\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n-1}\left(-{x}^{k-1}\right)\cdot xdx\)
\(=-kn{\int }_{0}^{1}{\left(1-{x}^{k}\right)}^{n-1}\left(1-{x}^{k}-1\right)dx\)
\({I}_{n}=-nk{\int }_{0}^{1}\left[{\left(1-{x}^{k}\right)}^{n}-{\left(1-{x}^{k}\right)}^{n-1}\right]dx\)
\({I}_{n}=nk{I}_{n-1}-nk{I}_{n}\)
\(\frac{{I}_{n}}{{I}_{n-1}}=\frac{nk}{nk+1}\)
\(\frac{{I}_{21}}{{I}_{20}}=\frac{21k}{1+21k}\)
\(=\frac{147}{148}\Rightarrow k=7\)
The integral \({\int }_{0}^{\pi }\frac{(\mathrm{x}+3)\mathrm{sinx}}{1+3{\cos }^{2}\mathrm{x}}\mathrm{dx}\) is equal to :
[JEE Main 2025, 7 Apr (Shift 1)]
\(\frac{\pi }{3\sqrt{3}}(\pi +6)\)
Sol: \(I={\int }_{0}^{\pi }\frac{(x+3)\sin x}{1+3{\cos }^{2}x}dx\)
\(I={\int }_{0}^{\pi \mathrm{/}2}\frac{(x+3)\sin x}{1+3{\cos }^{2}x}\\ +\frac{(p−x+3)\sin (\pi −x)}{1+3{\cos }^{2}(\pi −x)}dx\)
\(I={\int }_{0}^{\pi \mathrm{/}2}\frac{(\pi +6)\sin x}{1+3{\cos }^{2}x}dx\)
\(I−(\pi +6){\int }_{0}^{\pi \mathrm{/}2}\frac{\sin x}{1+3{\cos }^{2}x}\)
\(\begin{matrix}I=(\pi +6){\int }_{1}^{0}\frac{−dt}{1+3{t}^{2}}\ (∵\cos x=t) \\ =\frac{\pi +6}{3}{\int }_{0}^{1}\frac{dt}{{(\frac{1}{\sqrt{3}})}^{2}+{t}^{2}} \\ =\frac{\pi +6}{3}⋅\sqrt{3}⋅{({\tan }^{−1}\sqrt{3}t)}_{0}^{1} \\ =\frac{\pi +6}{\sqrt{3}}⋅\frac{\pi }{3}\end{matrix}\)
Let \(f:[0,\infty )\to \mathrm{ℝ}\)be differentiable function such that \(\mathrm{f}(\mathrm{x})=1-2\mathrm{x}+{\int }_{0}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}-\mathrm{t}}\mathrm{f}(\mathrm{t})\mathrm{dt}\) for all \(\mathrm{x}\in [0,\infty )\).Then the area of the region bounded by \(\mathrm{y}=f(\mathrm{x})\) and the coordinate axes is
[JEE Main 2025, 4 Apr (Shift 1)]
\(\frac{1}{2}\)
\(y=1−2x+{e}^{x}{\int }_{0}^{x}{e}^{−t}f(t)dt\)
\(\frac{dy}{dx}=−2+{e}^{−x}⋅{e}^{x}f(x)+{e}^{x}{\int }_{0}^{x}{e}^{−t}f(t)dt\)
\(\frac{dy}{dx}=−2+y+y+2x−1\)
\(\frac{dy}{dx}−2y=(2x−3)\)
\(I.F.={e}^{\int -2dx}={e}^{-2x}\)
solution is
\(y{e}^{−2x}=\int (2x−3)dx⋅{e}^{−2x}\)
\(y{e}^{−2x}=\frac{−(2x−3)}{2}{e}^{−2x}+\int {e}^{−2x}dx\)
\(y{e}^{−2x}=\frac{−(2x−3)}{2}{e}^{−2x}−\frac{1}{2}{e}^{−2x}+c\)
\(\mathrm{f}(0)=1\Rightarrow \mathrm{c}=1−\frac{3}{2}+\frac{1}{2}=0\)
\(y=−\frac{(2x−3)}{2}−\frac{1}{2}\)
\(y=-x+1\\ \Rightarrow x+y=1\)
Hence area bounded by \(y=f(x)\) and coordinate axes is
area \(=\frac{1}{2}(1)(1)=\frac{1}{2}\)
Let \(f(x)={\int }_{0}^{x}\left(t+\sin \left(1-{e}^{t}\right)\right)dt,x\in R\). Then, \(\lim _{x\to 0}\frac{f(x)}{{x}^{3}}\) is equal to
[JEE Main 2024, 4 Apr (Shift 2)]
\(-\frac{1}{6}\)
Let \(L=\lim_{x\to 0}\frac{f(x)}{x^3}\).
Since \(f(x)=\int_0^x\left(t+\sin(1-e^t)\right)dt\),
we have \(f(0)=0\).
Using L'Hospital's rule,
\(L=\lim_{x\to 0}\frac{x+\sin(1-e^x)}{3x^2}\).
Again it is of the form \( \frac{0}{0} \),
so applying L'Hospital's rule again,
\(L=\lim_{x\to 0}\frac{1-e^x\cos(1-e^x)}{6x}\).
Again it is of the form \( \frac{0}{0} \),
so applying L'Hospital's rule once more,
\(L=\lim_{x\to 0}\frac{-e^x\cos(1-e^x)-e^{2x}\sin(1-e^x)}{6}\).
Putting \(x=0\), we get
\(L=\frac{-1\cdot \cos 0-1\cdot \sin 0}{6}=-\frac{1}{6}\).
The integral \({\int }_{-1}^{\frac{3}{2}}\left(\left|{\pi }^{2}x\sin (\pi x)\right|\right)dx\) is equal to :
[JEE Main 2025, 8 Apr (Shift 1)]
\(1+3\pi\)
\(I={\pi }^{2}\int _{−1}^{3/2}|x\sin \pi x|dx\)
\(={\pi }^{2}\left\{\int _{−1}^{1}x\sin \pi xdx−\int _{1}^{3/2}x\sin \pi xdx\right\}\)
\(={\pi }^{2}\left\{2\int _{0}^{1}x\sin \pi xdx−\int _{−1}^{3/2}x\sin \pi xdx\right\}\)
\(\int x\sin \pi xdx−x⋅\frac{1}{\pi }\cos \pi x+\)\(\int 1⋅\frac{1}{\pi }\cos \pi xdx\)
\(={\pi }^{2}\left\{2\left(−\frac{x}{\pi }\cos \pi x+{\left.\frac{\sin \pi x}{{\pi }^{2}}\right)}_{0}^{1}\right.\right.\)
\(−\left(−\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^{2}}\right)\)
\(={\pi }^{2}\left\{\frac{2}{\pi }−\left(−\frac{1}{{\pi }^{2}}−\frac{1}{\pi }\right)\right\}\)
\(={\pi }^{2}\left\{{\left(−\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^{2}}\right)}_{0}^{1}\right.\)
\(−\left(−\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^{2}}\right)\)
\(={\pi }^{2}\left\{\frac{2}{\pi }−\left(−\frac{1}{{\pi }^{2}}−\frac{1}{\pi }\right)\right\}\)
\(={\pi }^{2}\left\{\frac{3}{\pi }+\frac{1}{{\pi }^{2}}\right\}\)
\(=3\pi +1\)
Let \(\mathrm{f}(\mathrm{x})=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, \mathrm{x} \in \mathrm{R}\), the number of local maximum and minimum point of \(f(x)\) respectively are (22 Jan, Shift II, Memory Based)
5
\(\begin{aligned}& \because f(x)=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} \\& f^{\prime}(x)=\frac{2 x\left(x^4-8 x^2+15\right)}{e^{x^2}} \\& =\frac{2 x\left(x^2-5\right)\left(x^2-3\right)}{e^{x^2}}\end{aligned}\)
The extremum value of \(f(x)\) are \(x=0, \pm \sqrt{5}, \pm \sqrt{3}\)
\(\therefore \quad\) Number of extremum points are 5.
The value of the integral \({\int }_{0}^{2}\frac{\sqrt{x\left({x}^{2}+x+1\right)}}{(\sqrt{x+1})\left(\sqrt{{x}^{4}+{x}^{2}+1}\right)}dx\) is equal to:
[JEE Main 2026, 8 Apr (Shift 2)]
\(\frac{2}{3}{\log }_{e}(3+2\sqrt{2})\)
Let \(I=\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^4+x^2+1}\right)}\right] d x\)
\(=\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1}) \cdot\left(\sqrt{\left(x^2+1\right)^2-x^2}\right)}\right] d x\)
\(=\int_0^2\left[\frac{\sqrt{x\left(x^2+x+1\right)}}{\left.(\sqrt{x+1}) \cdot\left\{\sqrt{\left(x^2+1+x\right)\left(x^2+1-x\right)}\right\}\right]}\right] d x\)
\(=\int_0^2\left[\frac{\sqrt{x}}{(\sqrt{x+1}) \cdot\left(\sqrt{x^2-x+1}\right)}\right] d x\)
\(=\int_0^2\left[\frac{\sqrt{x}}{\sqrt{x^3+1}}\right] d x\)
Let \(x=t^2 \Rightarrow d x=2 t d t\)
\(I=\int_0^{\sqrt{2}}\left[\frac{2 t^2}{\sqrt{t^6+1}}\right] d t=\left(\frac{2}{3}\right) \int_0^{\sqrt{2}}\left[\frac{3 t^2}{\sqrt{t^6+1}}\right] d t\)
Let \(t^3=u \Rightarrow 3 t^2 d t=d u\)
\(I=\left(\frac{2}{3}\right) \cdot \int_0^{2 \sqrt{2}}\left[\frac{d u}{\sqrt{u^2+1}}\right]\)
\(=\left(\frac{2}{3}\right) \cdot\left[\ln \left|u+\sqrt{u^2+1}\right|\right]_0^{2 \sqrt{2}}\)
\(=\left(\frac{2}{3}\right) \cdot[\ln (2 \sqrt{2}+3)-\ln 1]=\left(\frac{2}{3}\right) \cdot[\ln (3+2 \sqrt{2})]\)
Let for \(f(\mathrm{x})=7 \tan ^8 \mathrm{x}+7 \tan ^6 \mathrm{x}-3 \tan ^4 \mathrm{x}-3 \tan ^2 \mathrm{x}, \mathrm{I}_1=\int_0^{\pi / 4} f(\mathrm{x}) \mathrm{dx}\) and \(\mathrm{I}_2=\int_0^{\pi / 4} \mathrm{x} f(\mathrm{x}) \mathrm{dx}\). Then \(7 \mathrm{I}_1+12 \mathrm{I}_2\) is equal to :
[JEE Main 2025, 22 Jan (Shift 1)]
1
\(f(x)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x\)\(-3{\tan }^{2}x\)
\(f(x)=7{\tan }^{6}x\left({\tan }^{2}x+1\right)\)\(-3{\tan }^{2}x\left({\tan }^{2}x+1\right)\)
\(f(x)=\left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}x\)
On integrating both sides,we get
\({I}_{1}={\int }_{0}^{\pi /4}\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)dx\)
Put \(\tan x=t\Rightarrow dx={\sec }^{2}tdt\)
\({I}_{1}={\int }_{0}^{1}\left(7{t}^{6}-3{t}^{2}\right)dt={\left[{t}^{7}-{t}^{3}\right]}_{0}^{1}=0\\ {I}_{2}={\int }_{0}^{\pi /4}x\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)dx\\ ={\left[x\left({\tan }^{7}x-{\tan }^{3}x\right)\right]}_{0}^{\pi /4}-{\int }_{0}^{\pi /4}\left({\tan }^{7}x-{\tan }^{3}x\right)dx\)
\(=0-{\int }_{0}^{\pi /4}{\tan }^{3}x\left({\tan }^{2}x-1\right)\)\(\left(1+{\tan }^{2}x\right)dx\)
=\(-\int_0^{\pi / 4}\left(\tan ^5 x-\tan ^3 x\right) \sec ^2 x d x\)
Put \(\tan x=t \Rightarrow d x=\sec ^2 t d t\)
\(=-{\int }_{0}^{1}\left({t}^{5}-{t}^{3}\right)dt=-{\left[\frac{{t}^{6}}{6}-\frac{{t}^{4}}{4}\right]}_{0}^{1}=\frac{1}{12}\\ 7{\mathrm{I}}_{1}+12{\mathrm{I}}_{2}=1\)
The value of the integral \({\int }_{-1}^{2}{\log }_{e}\left(x+\sqrt{{x}^{2}+1}\right)dx\) is
[JEE Main 2024, 9 Apr (Shift 2)]
\(\sqrt{2}-\sqrt{5}+{\log }_{e}\left(\frac{9+4\sqrt{5}}{1+\sqrt{2}}\right)\)
Let \(I=\int_{-1}^{2}\log_e\left(x+\sqrt{x^2+1}\right)dx\).
Now \( \frac{d}{dx}\log_e\left(x+\sqrt{x^2+1}\right)=\frac{1}{\sqrt{x^2+1}} \).
So \(I=\left[x\log_e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^{2}-\int_{-1}^{2}\frac{x}{\sqrt{x^2+1}}dx\).
Since \( \int \frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1} \), we get
\(I=\left[x\log_e\left(x+\sqrt{x^2+1}\right)-\sqrt{x^2+1}\right]_{-1}^{2}\).
At \(x=2\), the value is \(2\log_e(2+\sqrt{5})-\sqrt{5}\).
At \(x=-1\), the value is \(-\log_e(\sqrt{2}-1)-\sqrt{2}\).
Therefore \(I=2\log_e(2+\sqrt{5})-\sqrt{5}+\log_e(\sqrt{2}-1)+\sqrt{2}\).
Now \(2\log_e(2+\sqrt{5})=\log_e(2+\sqrt{5})^2=\log_e(9+4\sqrt{5})\).
Also \( \sqrt{2}-1=\frac{1}{\sqrt{2}+1} \), so \( \log_e(\sqrt{2}-1)=-\log_e(\sqrt{2}+1) \).
Hence \(I=\sqrt{2}-\sqrt{5}+\log_e(9+4\sqrt{5})-\log_e(1+\sqrt{2})\).
So \(I=\sqrt{2}-\sqrt{5}+\log_e\left(\frac{9+4\sqrt{5}}{1+\sqrt{2}}\right)\).
If \(I={\int }_{0}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx,\) then \({\int }_{0}^{2I}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\) equals:
[JEE Main 2025, 23 Jan (Shift 2)]
\(\frac{{\pi }^{2}}{16}\)
For \(I={\int }_{0}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx\)
Apply king's rule
\(I={\int }_{0}^{\frac{\pi }{2}}\frac{{\cos }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx\)
Adding both we get:
\(2\mathrm{I}={\int }_{0}^{\pi /2}\mathrm{dx}=\frac{\pi }{2}\Rightarrow \mathrm{I}=\frac{\pi }{4}\)
\({I}_{2}={\int }_{0}^{\pi /2}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\)
Apply king's rule
\({I}_{2}={\int }_{0}^{\pi /2}\frac{\left(\frac{\pi }{2}-x\right)\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx\)
and adding both, we get
\(\text{ }{I}_{2}=\frac{\pi }{4}{\int }_{0}^{\pi /2}\frac{\tan x{\sec }^{2}xdx}{{\tan }^{4}x+1}\)
\(\text{ put }{\tan }^{2}\mathrm{x}=\mathrm{t}\Rightarrow \mathrm{tanx}{\sec }^{2}\mathrm{x}\mathrm{dx}=\frac{\mathrm{dt}}{2}\)
\({I}_{2}=\frac{\pi }{8}{\int }_{0}^{\infty }\frac{\mathrm{dt}}{{\mathrm{t}}^{2}+1}\)
\(=\frac{\pi }{8}{\left[{\tan }^{-1}\mathrm{t}\right]}_{0}^{\infty }\\ =\frac{\pi }{8}\cdot \frac{\pi }{2}=\frac{{\pi }^{2}}{16}\)
The value of the intergral \(\int_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}\) equals:
[JEE Main 2024, 1 Feb (Shift 1)]
\(\frac{\sqrt{2} \pi^2}{32}\)
Given integral is \(I=\int_0^{\pi/4}\frac{x\,dx}{\sin^4(2x)+\cos^4(2x)}\)
Let \(f(x)=\frac{1}{\sin^4(2x)+\cos^4(2x)}\)
Now, \(f\left(\frac{\pi}{4}-x\right)=\frac{1}{\sin^4\left(\frac{\pi}{2}-2x\right)+\cos^4\left(\frac{\pi}{2}-2x\right)}\)
So, \(f\left(\frac{\pi}{4}-x\right)=\frac{1}{\cos^4(2x)+\sin^4(2x)}=f(x)\)
Using the property, if \(f(a-x)=f(x)\), then \(\int_0^a xf(x)\,dx=\frac{a}{2}\int_0^a f(x)\,dx\)
Here \(a=\frac{\pi}{4}\)
Therefore, \(I=\frac{\pi}{8}\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}\)
Let \(J=\int_0^{\pi/4}\frac{dx}{\sin^4(2x)+\cos^4(2x)}\)
Put \(2x=t\)
Then \(2dx=dt\), so \(dx=\frac{dt}{2}\)
When \(x=0\), \(t=0\), and when \(x=\frac{\pi}{4}\), \(t=\frac{\pi}{2}\)
So, \(J=\frac{1}{2}\int_0^{\pi/2}\frac{dt}{\sin^4t+\cos^4t}\)
Now put \(\tan t=u\)
Then \(dt=\frac{du}{1+u^2}\)
Also, \(\sin^2t=\frac{u^2}{1+u^2}\) and \(\cos^2t=\frac{1}{1+u^2}\)
So, \(\sin^4t+\cos^4t=\frac{u^4+1}{(1+u^2)^2}\)
Therefore, \(\frac{dt}{\sin^4t+\cos^4t}=\frac{1+u^2}{1+u^4}\,du\)
When \(t=0\), \(u=0\), and when \(t=\frac{\pi}{2}\), \(u\to\infty\)
Thus, \(J=\frac{1}{2}\int_0^\infty \frac{1+u^2}{1+u^4}\,du\)
Now, \(u^4+1=(u^2+\sqrt2u+1)(u^2-\sqrt2u+1)\)
Also, \(\frac{1+u^2}{1+u^4}=\frac{1}{2}\left(\frac{1}{u^2+\sqrt2u+1}+\frac{1}{u^2-\sqrt2u+1}\right)\)
So, \(\int_0^\infty \frac{1+u^2}{1+u^4}\,du=\frac{\pi}{\sqrt2}\)
Therefore, \(J=\frac{1}{2}\cdot\frac{\pi}{\sqrt2}=\frac{\pi}{2\sqrt2}\)
Now, \(I=\frac{\pi}{8}\cdot J\)
\(I=\frac{\pi}{8}\cdot\frac{\pi}{2\sqrt2}\)
\(I=\frac{\pi^2}{16\sqrt2}\)
\(I=\frac{\sqrt2\pi^2}{32}\)
Hence, the value of the integral is \(\frac{\sqrt2\pi^2}{32}\)
The integral \({\int }_{0}^{1}{\cot }^{-1}\left(1+x+{x}^{2}\right)dx\) is equal to:
[JEE Main 2026, 4 Apr (Shift 2)]
\(2{\tan }^{-1}2-\frac{1}{2}{\log }_{\mathrm{e}}\left(\frac{5}{4}\right)-\frac{\pi }{2}\)
\( I=\int_0^1 \cot ^{-1}\left(1+x+x^2\right) d x\)
\(=\int_0^1 \tan ^{-1}\left(\frac{1}{1+x+x^2}\right) d x\)
\(=\int_0^1 \tan ^{-1}\left(\frac{x+1-x}{1+x(x+1)}\right) d x\)
\(I=\int_0^1 \tan ^{-1}(x+1) d x-\int_0^1 \tan ^{-1} x d x\)
\(=\int_1^2 \tan ^{-1} x d x-\int_0^1 \tan ^{-1} x d x\)
\(\because \int \tan ^{-1} x d x=x \tan ^{-1} x-\frac{1}{2} \ln \left(1+x^2\right)\)
\(\Rightarrow I=2 \tan ^{-1} 2-\frac{1}{2} \ln \left(\frac{5}{4}\right)-\frac{\pi}{2}\)
Let the domain of the function \(\mathrm{f}(\mathrm{x})={\log }_{2}{\log }_{4}{\log }_{6}\left(3+4\mathrm{x}-{\mathrm{x}}^{2}\right)\) be \((a,b)\). If \({\int }_{0}^{\mathrm{b}-\mathrm{a}}\left[{\mathrm{x}}^{2}\right]\mathrm{dx}\) \(=\mathrm{p}-\sqrt{\mathrm{q}}-\sqrt{\mathrm{r}},\mathrm{p},\mathrm{q},\mathrm{r}\in \mathrm{ℕ},\gcd (\mathrm{p},\mathrm{q},\mathrm{r})=1\), where \([\cdot ]\) is the greatest integer function, then \(\mathrm{p}+\mathrm{q}+\mathrm{r}\) is equal to
[JEE Main 2025, 3 Apr (Shift 1)]
\(10\)
\(f(x)={\log }_{2}{\log }_{4}{\log }_{6}(3+4x−{x}^{2})\)
\({\log }_{6}(3+4x−{x}^{2})>0\Rightarrow 3+4x−{x}^{2}>1\)
\(−{x}^{2}+4x+2>0\Rightarrow {x}^{2}−4x−2<0\)
Roots: \(x=2\pm \sqrt{6}\).
So, \(2−\sqrt{6} \({\log }_{4}({\log }_{6}(3+4x−{x}^{2}))>0\)\(\Rightarrow {\log }_{6}(3+4x−{x}^{2})>1\) \(3+4x−{x}^{2}>6\Rightarrow −{x}^{2}+4x−3>0\)\(\Rightarrow {x}^{2}−4x+3<0\) so \(1 \({\log }_{2}({\log }_{4}{\log }_{6}(⋅))\text{ requires }{\log }_{4}{\log }_{6}(⋅)>0,\) which is the same as step 2. So \(\text{domain}=(a,b)=(1,3)\mathrm{.}\) Thus \(b−a=2\) and we need \({\int }_{0}^{2}⌊{x}^{2}⌋\text{ }dx\mathrm{.}\) On \([0,2]\), \({x}^{2}\) runs from 0 to 4; break at where \({x}^{2}=k\): So \({\int }_{0}^{2}⌊{x}^{2}⌋dx={\int }_{0}^{1}0\text{ }dx+{\int }_{1}^{\sqrt{2}}1\text{ }dx\)\(+{\int }_{\sqrt{2}}^{\sqrt{3}}2\text{ }dx+{\int }_{\sqrt{3}}^{2}3\text{ }dx\) \(=(\sqrt{2}−1)+2(\sqrt{3}−\sqrt{2})+3(2−\sqrt{3})\mathrm{.}\) Simplify: \(=\sqrt{2}−1+2\sqrt{3}−2\sqrt{2}+6−3\sqrt{3}\)\(=5−\sqrt{2}−\sqrt{3}\mathrm{.}\) Thus it matches \(p−\sqrt{q}−\sqrt{r}\) with \(p+q+r=5+2+3=10.\)
If \(\alpha =1\) and \(\beta =1+i\sqrt{2}\), where \(i=\sqrt{-1}\) are two roots of the equation \({x}^{3}+a{x}^{2}+bx+c=0,a,b,c\in \mathrm{R}\), then \({\int }_{-1}^{1}\left({x}^{3}+a{x}^{2}+bx+c\right)dx\) is equal to:
[JEE Main 2026, 4 Apr (Shift 2)]
\(–8\)
Roots are \(1 \pm \mathrm{i} \sqrt{2} \) and \(1\)
Sum of roots:
\(-\mathrm{a}=3 \Rightarrow \mathrm{a}=-3\)
Product of roots:
\(-\mathrm{c}=1(1+\sqrt{2} \mathrm{i})(1-\mathrm{i} \sqrt{2})\)
\(\Rightarrow \mathrm{c}=-3\)
\(I=\int_{-1}^1\left(x^3-3 x^2+b x-3\right) d x\)
\(=2 \int_0^1\left(-3 x^2-3\right) d x\)
\(=2\left(-x^3-3 x\right)_0^1=-8\)
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
\(\frac{3 \pi^{2}}{8}\)
\(L=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)\)
Applying L-Hospital's Rule
\(\Rightarrow L=\lim _{x\to \frac{\pi }{2}}\frac{−\cos (x)3{x}^{2}}{2\left(x−\frac{\pi }{2}\right)}=\lim _{x\to \frac{\pi }{2}}\frac{\sin \left(x−\frac{\pi }{2}\right)3{x}^{2}}{2\left(x−\frac{\pi }{2}\right)}\)
\(\Rightarrow L=\frac{3{\pi }^{2}}{8}\)
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)\) is equal to
[JEE Main 2024, 29 Jan (Shift 1)]
\(\frac{3 \pi^{2}}{8}\)
\(L=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)\)
Applying L-Hospital's Rule
\(\Rightarrow L=\lim _{x\to \frac{\pi }{2}}\frac{−\cos (x)3{x}^{2}}{2\left(x−\frac{\pi }{2}\right)}=\lim _{x\to \frac{\pi }{2}}\frac{\sin \left(x−\frac{\pi }{2}\right)3{x}^{2}}{2\left(x−\frac{\pi }{2}\right)}\)
\(\Rightarrow L=\frac{3{\pi }^{2}}{8}\)
The integral \(80{\int }_{0}^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta }\right)d\theta\) is equal to :
[JEE Main 2025, 29 Jan (Shift 1)]
\(4 \log _e 3\)
\(\text{Let}I=80{\int }_{0}^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9+16(2\sin \theta \cdot \cos \theta )}\right)d\theta \\ =80{\int }_{0}^{\frac{\pi }{4}}\frac{\sin \theta +\cos \theta }{9+16-16(\sin \theta -\cos \theta {)}^{2}}d\theta \\ \text{Put}\sin \theta -\cos \theta =t\\ \left(\cos \theta +\sin \theta \right)\mathrm{d}\theta =dt\\ I=80{\int }_{-1}^{0}\frac{\mathrm{dt}}{25-16{t}^{2}}\\ =\frac{80}{16}{\int }_{-1}^{0}\frac{\mathrm{dt}}{{\left(\frac{5}{4}\right)}^{2}-{\mathrm{t}}^{2}}{\left.=\frac{5}{2\left(\frac{5}{4}\right)}\ln \left(\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right)\right]}_{-1}^{0}\\ =4\ln 3\)
\(\text { If } I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \text {, then the value of definite integration } \int_0^{2 I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \text { is }\)
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Let \(\left(2^{1-\mathrm{a}} + 2^{1+\mathrm{a}}\right), f(a),\left(3^{\mathrm{a}}+3^{-\mathrm{a}}\right)\) be in A.P. and \(\alpha\) be the minimum value of \(f (a)\). Then the value of the integral \(\int_{\log _e(\alpha-1)}^{\log _e(\alpha)} \frac{d x}{\left(e^{2 x}-e^{-2 x}\right)}\) is:
[JEE Main 2026, 5 Apr (Shift 2)]
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Let \((a,b)\) be the point of intersection of the curve \({x}^{2}=2y\) and the straight line \(y-2x-6=0\) in the second quadrant. Then the integral \(I={\int }_{a}^{b}\frac{9{x}^{2}}{1+{5}^{x}}dx\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
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Let \(\mathrm{f}:[1,\infty )\to [2,\infty )\) be a differentiable function, If \(10{\int }_{1}^{x}f(\mathrm{t})\mathrm{dt}=5\mathrm{x}f(\mathrm{x})-{\mathrm{x}}^{5}-9\) for all \(\mathrm{x}\geq 1\), then the value of \(f(3)\) is :
[JEE Main 2025, 2 Apr (Shift 2)]
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Let \(f(x)\) be a positive function and \(I_1=\int_{-\frac{1}{2}}^1 2 x f(2 x(1-2 x)) d x\) and \(I_2=\int_{-1}^2 f(x(1-x)) d x\). Then the value of \(\frac{I_2}{I_1}\) is equal to
[JEE Main 2025, 8 Apr (Shift 1)]
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The value of \({\int }_{{\mathrm{e}}^{2}}^{{\mathrm{e}}^{4}}\frac{1}{\mathrm{x}}\left(\frac{{\mathrm{e}}^{{\left({\left({\log }_{\mathrm{e}}\mathrm{x}\right)}^{2}+1\right)}^{-1}}}{{\mathrm{e}}^{{\left({\left({\log }_{e}x\right)}^{2}+1\right)}^{-1}}+{\mathrm{e}}^{{\left({\left(6-{\log }_{e}x\right)}^{2}+1\right)}^{-1}}}\right)dx\) is
[JEE Main 2025, 23 Jan (Shift 1)]
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\(\text{ If }{\int }_{-\pi /2}^{\pi /2}\frac{96\left({x}^{2}+\cos x\right)}{1+{e}^{x}}dx=\alpha {\pi }^{3}+\beta \text{ (where }\alpha ,\beta \text{ are positive integers),}\\ \text{then }\alpha +\beta \text{ equal to }\) (28 Jan, Shift I, Memory Based)
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Let \(f\) be a real valued continuous function defined on the positive real axis such that \(g(x)=\int_0^x t f(t) d t\)
If \(\mathrm{g}\left(x^3\right)=x^6+x^7\), then value of \(\sum_{r=1}^{15} f\left(\mathrm{r}^3\right)\) is :
[JEE Main 2025, 28 Jan (Shift 2)]
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If \(\mathrm{I}(\mathrm{~m}, \mathrm{n})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx}, \mathrm{~m}, \mathrm{n}>0 \) then \( \mathrm{I}(9,14)+\mathrm{I}(10,13),\) is
[JEE Main 2025, 24 Jan (Shift 1)]
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\(\text { If } I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \text {, then the value of definite integration } \int_0^{2 I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \text { is }\)
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Let the domain of the function \(\mathrm{f}(\mathrm{x})={\log }_{2}{\log }_{4}{\log }_{6}\left(3+4\mathrm{x}-{\mathrm{x}}^{2}\right)\) be \((a,b)\). If \({\int }_{0}^{\mathrm{b}-\mathrm{a}}\left[{\mathrm{x}}^{2}\right]\mathrm{dx}\) \(=\mathrm{p}-\sqrt{\mathrm{q}}-\sqrt{\mathrm{r}},\mathrm{p},\mathrm{q},\mathrm{r}\in \mathrm{ℕ},\gcd (\mathrm{p},\mathrm{q},\mathrm{r})=1\), where \([\cdot ]\) is the greatest integer function, then \(\mathrm{p}+\mathrm{q}+\mathrm{r}\) is equal to
[JEE Main 2025, 3 Apr (Shift 1)]
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The integral \({\int }_{0}^{\frac{\pi }{4}}\frac{136\sin x}{3\sin x+5\cos x}dx\) is equal to :
[JEE Main 2024, 5 Apr (Shift 1)]
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The value of \({\int }_{-\pi }^{\pi }\frac{2y(1+\sin y)}{1+{\cos }^{2}y}dy\) is:
[JEE Main 2024, 5 Apr (Shift 1)]
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If \({\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96{x}^{2}{\cos }^{2}x}{\left(1+{e}^{x}\right)}dx=\pi \left(\alpha {\pi }^{2}+\beta \right),\alpha ,\beta \in Z,\) then \((\alpha +\beta {)}^{2}\) equals
[JEE Main 2025, 28 Jan (Shift 1)]
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The value of the integral \({\int }_{-1}^{1}\left(\frac{{x}^{3}+|x|+1}{{x}^{2}+2|x|+1}\right)dx\) is equal to:
[JEE Main 2026, 6 Apr (Shift 2)]
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Evaluate: \(I=80{\int }_{0}^{\frac{\pi }{2}}\frac{\sin x+\cos x}{9\sin x+16\cos x}dx\)
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Let \(f\) be a polynomial function such that \(f\left({x}^{2}+1\right)={x}^{4}+5{x}^{2}+2\), for all \(x\in ℝ\). Then \({\int }_{0}^{3}f\left(x\right)dx\) is equal to
[JEE Main 2026, 28 Jan (Shift 1)]
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The value of \(\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}\) is:
[JEE Main 2024, 30 Jan (Shift 1)]
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The value of \(\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}\) is:
[JEE Main 2024, 30 Jan (Shift 1)]
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Let \(f(x)=\int_0^x \mathrm{t}\left(\mathrm{t}^2-9 \mathrm{t}+20\right) \mathrm{dt}, 1 \leq x \leq 5\). If the range of \(f\) is \([\alpha, \beta]\), then \(4(\alpha+\beta)\) equals :
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For \(0<\mathrm{a}<1\), the value of the integral \(\int_{0}^{\pi} \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^{2}}\) is:
[JEE Main 2024, 27 Jan (Shift 2)]
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For \(0<\mathrm{a}<1\), the value of the integral \(\int_{0}^{\pi} \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^{2}}\) is:
[JEE Main 2024, 27 Jan (Shift 2)]
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\(If{\int }_{0}^{x}t.f(t)dt={x}^{2}f(x),andf(2)=3,thenf(6)isequalsto\) (28 Jan, Shift I, Memory Based)
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The value of \({\int }_{−\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\frac{1}{\left[x\right]+4}\right)dx\), where \(\left[\cdot \right]\) denotes the greatest integer function, is
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Let \(\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}\) be a twice differentiable function such that \(f(2)=1\). If \(\mathrm{F}(x)=x f(x)\) for all \(x \in \mathbf{R}\), \(\int_0^2 x \mathrm{~F}^{\prime}(x) \mathrm{d} x=6\) and \(\int_0^2 x^2 \mathrm{~F}^{\prime \prime}(x) \mathrm{d} x=40\), then \(\mathrm{F}^{\prime}(2)+\int_0^2 \mathrm{~F}(x) \mathrm{d} x\) is equal to :
[JEE Main 2025, 28 Jan (Shift 2)]
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If \((I={\int }_{0}^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}x}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}dx),\) then \(({\int }_{0}^{2I}\frac{x\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx)\) equals:
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\(\text { If } I(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0 \text {, then } I(9,14)+I(10,13) \text { is equal to }\) (24 Jan, Shift I, Memory Based)
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If the value of the integral \({\int }_{-1}^{1}\frac{\cos \alpha x}{1+{3}^{x}}dx\) is \(\frac{2}{\pi }\). Then, a value of \(\alpha\) is
[JEE Main 2024, 4 Apr (Shift 2)]
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Let for some function \(\mathrm{y}=\mathrm{f}(\mathrm{x}),{\int }_{0}^{\mathrm{x}}\mathrm{tf}(\mathrm{t})\mathrm{dt}={\mathrm{x}}^{2}\mathrm{f}(\mathrm{x}),\) \(x>0\text{ and }f(2)=3.\) Then \(f(6)\) is equal to :
[JEE Main 2025, 28 Jan (Shift 1)]
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Let \(f(x)={\int }_{0}^{x}t\left({t}^{2}-3t+20\right)dt,x\in (1,3)\) and range of f(x) is \((\alpha ,\beta )\), then \(\alpha +\beta\) is equal to
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If \({\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96{x}^{2}{\cos }^{2}x}{\left(1+{e}^{x}\right)}dx=\pi \left(\alpha {\pi }^{2}+\beta \right),\alpha ,\beta \in Z,\) then \((\alpha +\beta {)}^{2}\) equals
[JEE Main 2025, 28 Jan (Shift 1)]
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The integral \({\int }_{0}^{\pi }\frac{8xdx}{4{\cos }^{2}x+{\sin }^{2}x}\) is equal to
[JEE Main 2025, 3 Apr (Shift 2)]
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\(4{\int }_{0}^{1}\left(\frac{1}{\sqrt{3+{\mathrm{x}}^{2}}+\sqrt{1+{\mathrm{x}}^{2}}}\right)\mathrm{dx}-3{\log }_{\mathrm{e}}(\sqrt{3})\) is equal to :
[JEE Main 2025, 2 Apr (Shift 2)]
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If \( \int_0^{\frac{\pi}{3}} \cos ^4 x d x= a \pi+ b \sqrt{3} \) where \(a\) and \(b\) are rational numbers, then \(9 a +8 b\) is equal to : EndFragment
[JEE Main 2024, 01 Feb (Shift 2)]
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If \( \int_0^{\frac{\pi}{3}} \cos ^4 x d x= a \pi+ b \sqrt{3} \) where \(a\) and \(b\) are rational numbers, then \(9 a +8 b\) is equal to : EndFragment
[JEE Main 2024, 01 Feb (Shift 2)]
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Let \(\int_{-2}^2(|\sin x|+[x \sin x]) d x=2(3-\cos 2)+\beta\) where \([\cdot ]\) is the greatest integer function. Then \(\beta \sin \left(\frac{\beta }{2}\right)\)equals:
[JEE Main 2026, 4 Apr (Shift 1)]
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Let \(f(x)=\left\{\begin{array}{ll}\frac{1}{3}, & x \leq \frac{\pi} { 2} \\ \frac{b(1-\sin x)}{(\pi-2 x)^2}, & x>\frac{\pi} { 2}\end{array}\right.\)
If \(f\) is continuous at \(x=\frac{\pi} { 2}\), then the value of \(\int_0^{3 b-6}\left|x^2+2 x-3\right| d x\) is:
[JEE Main 2026, 8 Apr (Shift 2)]
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The value of the integral \({\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\left(\frac{32{\cos }^{4}x}{1+{e}^{\sin x}}\right)dx\) is:
[JEE Main 2026, 6 Apr (Shift 1)]
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Let \(\mathrm{f}:[1,\infty )\to [2,\infty )\) be a differentiable function, If \(10{\int }_{1}^{x}f(\mathrm{t})\mathrm{dt}=5\mathrm{x}f(\mathrm{x})-{\mathrm{x}}^{5}-9\) for all \(\mathrm{x}\geq 1\), then the value of \(f(3)\) is :
[JEE Main 2025, 2 Apr (Shift 2)]
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\(\text{ If }{\int }_{-\pi /2}^{\pi /2}\frac{96\left({x}^{2}+\cos x\right)}{1+{e}^{x}}dx=\alpha {\pi }^{3}+\beta \text{ (where }\alpha ,\beta \text{ are positive integers),}\\ \text{then }\alpha +\beta \text{ equal to }\) (28 Jan, Shift I, Memory Based)
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Let \(\mathrm{f}(\mathrm{x})=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, \mathrm{x} \in \mathrm{R}\), the number of local maximum and minimum point of \(f(x)\) respectively are (22 Jan, Shift II, Memory Based)
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Let \(A=\left[\begin{matrix}1 & 3 & -1 \\ 2 & 1 & \alpha \\ 0 & 1 & -1\end{matrix}\right]\) be a singular matrix. Let \(f(x)={\int }_{0}^{x}\left({t}^{2}+2t+3\right)dt,x\in [1,\alpha ]\). If \(M\) and \(m\) are respectively be the maximum and the minimum values of \(f\) in \([1,\alpha ]\), then \(3(M-m)\) is equal to:
[JEE Main 2026, 6 Apr (Shift 2)]
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The value of \({\int }_{-1}^{1}\frac{(1+\sqrt{|x|-x}){e}^{x}+(\sqrt{|x|-x}){e}^{-x}}{{e}^{x}+{e}^{-x}}dx\) is equal to
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Evaluate: \(I=80{\int }_{0}^{\frac{\pi }{2}}\frac{\sin x+\cos x}{9\sin x+16\cos x}dx\)
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Let \(\beta(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0\). If \(\int_0^1\left(1-x^{10}\right)^{20} d x=a \times \beta(b, c)\), then \(100(a+b+c)\) equals
[JEE Main 2024, 5 Apr (Shift 2)]
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Let \(f:R\to R\) be such that \(f(xy)=f(x)f(y)\), for all \(x,y\in R\) and \(f(0)\neq 0\). Let \(g:[1,\infty )\to R\) be a differentiable function such that \({x}^{2}g(x)={\int }_{1}^{x}\left({t}^{2}f(t)-tg(t)\right)dt\). Then \(g(2)\) is equal to:
[JEE Main 2026, 6 Apr (Shift 2)]
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The integral \(80{\int }_{0}^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta }\right)d\theta\) is equal to :
[JEE Main 2025, 29 Jan (Shift 1)]
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Let \(a\) and \(b\) be real constants such that the function \(f\) defined by \(f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.\) be differentiable on \(R\). Then, the value of \(\int_{-2}^2 f(x) d x\) equals
[JEE Main 2024, 30 Jan (Shift 2)]
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Let \(a\) and \(b\) be real constants such that the function \(f\) defined by \(f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.\) be differentiable on \(R\). Then, the value of \(\int_{-2}^2 f(x) d x\) equals
[JEE Main 2024, 30 Jan (Shift 2)]
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The integral \({\int }_{-1}^{\frac{3}{2}}\left(\left|{\pi }^{2}x\sin (\pi x)\right|\right)dx\) is equal to :
[JEE Main 2025, 8 Apr (Shift 1)]
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Let \(f:[1, \infty) \rightarrow \mathbf{R}\) be a differentiable function defined as \(f(x)=\int_1^x f(\mathrm{t}) \mathrm{dt}+(1-x)\left(\log _{\mathrm{e}} x-1\right)+\mathrm{e}\). Then the value of \(f(f(1))\) is:
[JEE Main 2026, 5 Apr (Shift 2)]
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Let \(f(x)=\int_0^x \mathrm{t}\left(\mathrm{t}^2-9 \mathrm{t}+20\right) \mathrm{dt}, 1 \leq x \leq 5\). If the range of \(f\) is \([\alpha, \beta]\), then \(4(\alpha+\beta)\) equals :
[JEE Main 2025, 29 Jan (Shift 2)]
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Let \(f(x)={\int }_{0}^{x}t\left({t}^{2}-3t+20\right)dt,x\in (1,3)\) and range of f(x) is \((\alpha ,\beta )\), then \(\alpha +\beta\) is equal to
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\(\text { If } I(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0 \text {, then } I(9,14)+I(10,13) \text { is equal to }\) (24 Jan, Shift I, Memory Based)
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Let \(f\left(t\right)=\int \left(\frac{1−\sin \left({\log }_{e}t\right)}{1−\cos \left({\log }_{e}t\right)}\right)dt,t>1.\) If \(f\left({e}^{\pi /2}\right)=−{e}^{\pi /2}\) and \(f\left({e}^{\pi /4}\right)=\alpha {e}^{\pi /4},\) then \(\alpha\) equals
[JEE Main 2026, 24 Jan (Shift 1)]
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Let for \(f(\mathrm{x})=7{\tan }^{8}\mathrm{x}+7{\tan }^{6}\mathrm{x}-3{\tan }^{4}\mathrm{x}-3{\tan }^{2}\mathrm{x},\) \({\mathrm{I}}_{1}={\int }_{0}^{\pi /4}f(\mathrm{x})\mathrm{dx}\text{ and }\)\({\mathrm{I}}_{2}={\int }_{0}^{\pi /4}\mathrm{x}f(\mathrm{x})\mathrm{dx}.\mathrm{Then}\) \(7{\mathrm{I}}_{1}+12{\mathrm{I}}_{2}\) is equal to :
[JEE Main 2025, 22 Jan (Shift 1)]
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The value of the integral \({\int }_{0}^{\infty }\frac{{\log }_{e}(x)}{{x}^{2}+4}dx\) is:
[JEE Main 2026, 5 Apr (Shift 1)]
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Let \((a,b)\) be the point of intersection of the curve \({x}^{2}=2y\) and the straight line \(y-2x-6=0\) in the second quadrant. Then the integral \(I={\int }_{a}^{b}\frac{9{x}^{2}}{1+{5}^{x}}dx\) is equal to :
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The integral \({\int }_{0}^{\pi }\frac{(\mathrm{x}+3)\mathrm{sinx}}{1+3{\cos }^{2}\mathrm{x}}\mathrm{dx}\) is equal to :
[JEE Main 2025, 7 Apr (Shift 1)]
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The value of \({\int }_{0}^{\frac{\pi }{4}}\left(\sin \left|\left(4x-\frac{\pi }{2}\right)\right|+\sin [2x]\right)dx\) is (where [•] denotes the greatest integer function)
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\({\int }_{1}^{2}\frac{\cos (\log x)}{x}\text{ }dx=\)
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Let \(f:[0,\infty )\to \mathrm{ℝ}\)be differentiable function such that \(\mathrm{f}(\mathrm{x})=1-2\mathrm{x}+{\int }_{0}^{\mathrm{x}}{\mathrm{e}}^{\mathrm{x}-\mathrm{t}}\mathrm{f}(\mathrm{t})\mathrm{dt}\) for all \(\mathrm{x}\in [0,\infty )\).
Then the area of the region bounded by \(\mathrm{y}=f(\mathrm{x})\) and the coordinate axes is
[JEE Main 2025, 4 Apr (Shift 1)]
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Let \([x]\) denote the greatest integer function. Then \({\int }_{−\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\frac{12\text{ }\left(3+\left[x\right]\right)}{3\text{ }+\text{ }\left[\sin \text{ }x\right]\text{ }+\text{ }\left[\cos \text{ }x\right]}\right)\text{ }dx\) is equal to:
[JEE Main 2026, 28 Jan (Shift 2)]
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If \(\mathrm{I}(\mathrm{m},\mathrm{n})={\int }_{0}^{1}{\mathrm{x}}^{\mathrm{m}-1}(1-\mathrm{x}{)}^{\mathrm{n}-1}\mathrm{dx},\mathrm{m},\mathrm{n}>0\), then \(\mathrm{I}(9,14)+\mathrm{I}(10,13)\) is
[JEE Main 2025, 24 Jan (Shift 1)]
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The value of \({\int }_{{\mathrm{e}}^{2}}^{{\mathrm{e}}^{4}}\frac{1}{\mathrm{x}}\left(\frac{{\mathrm{e}}^{{\left({\left({\log }_{\mathrm{e}}\mathrm{x}\right)}^{2}+1\right)}^{-1}}}{{\mathrm{e}}^{{\left({\left({\log }_{e}x\right)}^{2}+1\right)}^{-1}}+{\mathrm{e}}^{{\left({\left(6-{\log }_{e}x\right)}^{2}+1\right)}^{-1}}}\right)dx\) is
[JEE Main 2025, 23 Jan (Shift 1)]
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\(If{\int }_{0}^{x}t.f(t)dt={x}^{2}f(x),andf(2)=3,thenf(6)isequalsto\) (28 Jan, Shift I, Memory Based)
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The value of \({\int }_{0}^{\frac{\pi }{4}}\left(\sin \left|\left(4x-\frac{\pi }{2}\right)\right|+\sin [2x]\right)dx\) is (where [•] denotes the greatest integer function)
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The value of \({\int }_{0}^{20\pi }\left({\sin }^{4}x+{\cos }^{4}x\right)dx\) is equal to:
[JEE Main 2026, 2 Apr (Shift 2)]
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Let \(f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R\) be a differentiable function such that \(f(0)=\frac{1}{2}\). If the \(\lim _{x \rightarrow 0} \frac{x \int_0^x f( t ) dt }{ e ^{x^2}-1}=\alpha\), then \(8 \alpha^2\) is equal to :
[JEE Main 2024, 30 Jan (Shift 1)]
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Let \(f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R\) be a differentiable function such that \(f(0)=\frac{1}{2}\). If the \(\lim _{x \rightarrow 0} \frac{x \int_0^x f( t ) dt }{ e ^{x^2}-1}=\alpha\), then \(8 \alpha^2\) is equal to :
[JEE Main 2024, 30 Jan (Shift 1)]
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Let \(f\) be a real valued continuous function defined on the positive real axis such that \(g(x)=\int_0^x t f(t) d t\). If \(\mathrm{g}\left(x^3\right)=x^6+x^7\), then value of \(\sum_{r=1}^{15} f\left(\mathrm{r}^3\right)\) is :
[JEE Main 2025, 28 Jan (Shift 2)]
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The integral\({\int }_{1/4}^{3/4}\cos \left(2{\cot }^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx\) is equal to
[JEE Main 2024, 9 Apr (Shift 2)]
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Let \(f:R\to R\) be defined as
\(f\left(x\right)=a{e}^{2x}+b{e}^{x}+cx.\)
If \(f\left(0\right)=-1,f'\left({\log }_{e}2\right)=21\) and
\({\int }_{0}^{{\log }_{e}4}\left(f\left(x\right)-cx\right)dx=\frac{39}{2},\) then the
value of \(\left|a+b+c\right|\) equals
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Let \(f:R\to R\) be defined as
\(f\left(x\right)=a{e}^{2x}+b{e}^{x}+cx.\)
If \(f\left(0\right)=-1,f'\left({\log }_{e}2\right)=21\) and
\({\int }_{0}^{{\log }_{e}4}\left(f\left(x\right)-cx\right)dx=\frac{39}{2},\) then the
value of \(\left|a+b+c\right|\) equals
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If the value of the integral \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x=\frac{\pi}{4}(\pi+a)-2\), then the value of \(a\) is
[JEE Main 2024, 29 Jan (Shift 1)]
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If the value of the integral \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x=\frac{\pi}{4}(\pi+a)-2\), then the value of \(a\) is
[JEE Main 2024, 29 Jan (Shift 1)]
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The value of \(\int_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} d x \) is equal to :
[JEE Main 2024, 1 Feb (Shift 2)]
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The integral \({\int }_{0}^{\pi }\frac{8xdx}{4{\cos }^{2}x+{\sin }^{2}x}\) is equal to
[JEE Main 2025, 3 Apr (Shift 2)]
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The value of the integral \({\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\left(\frac{4-{\csc }^{2}x}{{\cos }^{4}x}\right)dx\) is:
[JEE Main 2026, 5 Apr (Shift 1)]
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Let \([\cdot ]\) denote the greatest integer function. Then the value of \({\int }_{0}^{3}\left(\frac{{e}^{x}+{e}^{-x}}{\left[|x|\right]!}\right)dx\) is:
[JEE Main 2026, 2 Apr (Shift 1)]
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The value of \({\int }_{−\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\frac{1}{\left[x\right]+4}\right)dx\), where \(\left[\cdot \right]\) denotes the greatest integer function, is
[JEE Main 2026, 22 Jan (Shift 1)]
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The value of the integral \({\int }_{1}^{2}\left(\frac{{t}^{4}+1}{{t}^{6}+1}\right)dt\) is:
[JEE Main 2023, 29 Jan (Shift 2)]
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The value of \({\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{1}{1+{e}^{\sin x}}dx\) is
[JEE Main 2020, 5 Sep (Shift 1)]
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If \({\int }_{0}^{100\pi }\frac{{\sin }^{2}x}{{e}^{\left(\frac{x}{\pi }-\left[\frac{x}{\pi }\right]\right)}}dx=\frac{a{\pi }^{3}}{1+4{\pi }^{2}},a\in R\), where \([x]\) is the greatest integer less than or equal to \(x\), then the value of \(\alpha\) is:
[JEE Main 2021, 22 Jul (Shift 2)]
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The integral \(16{\int }_{1}^{2}\frac{dx}{{x}^{3}{\left({x}^{2}+2\right)}^{2}}\) is equal to
[JEE Main 2023, 25 Jan (Shift 2)]
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The value of the integral \({\int }_{1/2}^{2}\frac{{\tan }^{-1}x}{x}dx\) is equal to
[JEE Main 2023, 29 Jan (Shift 2)]
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The value of the integral \({\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{x+\frac{\pi }{4}}{2-\cos 2x}dx\) is:
[JEE Main 2023, 1 Feb (Shift 2)]
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If \([x]\) denotes the greatest integer less than or equal to \(x\), then the value of the integral \(\int_{-\pi / 2}^{\pi / 2}[[x]-\sin x] d x\) is equal to:
[JEE Main 2021, 20 Jul (Shift 2)]
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If \([x]\) is the greatest integer \(\leq x\), then \(\pi^2 \int_0^2\left(\sin \frac{\pi x}{2}\right)(x-[x])^{[x]} d x\) is equal to:
[JEE Main 2021, 31 Aug (Shift 2)]
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The value of \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^n \frac{(2 j-1)+8 n}{(2 j-1)+4 n}\) is equal to:
[JEE Main 2021, 27 Jul (Shift 1)]
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The value of the integer \({\int }_{0}^{1}\frac{\sqrt{x}dx}{(1+x)(1+3x)(3+x)}\) is:
[JEE Main 2021, 27 Aug (Shift 2)]
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The value of the definite integral \({\int }_{\pi /24}^{5\pi /24}\frac{dx}{1+\sqrt[3]{\tan 2x}}\)
[JEE Main 2021, 25 Jul (Shift 1)]
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\(\int_0^1 \frac{1}{\left(5+2 x-2 x^2\right)\left(1+e^{(2-4 x)}\right)} d x=\frac{1}{\alpha} \log _e\left(\frac{\alpha+1}{\beta}\right)\).
\(\alpha, \beta,>0\), then \(\alpha^4-\beta^4\) is equal to:
[JEE Main 2023, 15 Apr (Shift 1)]
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The minimum value of the function \(f(x)=\int_0^2 e^{|x-t|} d t\) is
[JEE Main 2023, 25 Jan (Shift 1)]
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The value of \({\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\frac{1+{\sin }^{2}x}{1+{\pi }^{\sin x}}\right)dx\) is:
[JEE Main 2021, 26 Aug (Shift 2)]
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The value of \(\sum_{n=1}^{100} \int_{n-1}^n e^{x-[x]} d x\), where \([x]\) is the greatest integer \(\leq x\), is:
[JEE Main 2021, 26 Feb (Shift 1)]
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\(\lim _{n \rightarrow \infty}\left(\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots+\frac{1}{2 n}\right)\) is equal to
[JEE Main 2023, 1 Feb (Shift 1)]
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Let \(f:(0,1) \rightarrow R\) be the functions defined as \(f(x)=\sqrt{n}\) if \(x \in\left[\frac{1}{n+1}, \frac{1}{n}\right)\) where \(n \in N\). Let \(g :(0,1) \rightarrow R\) be a function such that \(\int_{x^2}^x \sqrt{\frac{1-t}{t}}d t
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If \(I_{m, n}=\int_0^1 x^{m-1}(1-x)^{n-1} d x\), for \(m, n \geq 1\), and \(\int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} d x=\alpha I_{m, n}\), \(\alpha \in R\) then \(\alpha\) equals _________.
[JEE Main 2021, 26 Feb (Shift 2)]
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The value of \( \int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{3} x} \) is
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Let \(f\) be a differentiable function defined on \(\left[0, \frac{\pi}{2}\right]\) such that \(f(x)>0\) and \(f(x)+\int_0^x f(t) \sqrt{1-\left(\log _e f(t)\right)^2} d t=e, \forall x \in\left[0, \frac{\pi}{2}\right]\) then \(\left(6 \log _e f\left(\frac{\pi}{6}\right)\right)^2\) is equal to ___________.
[JEE Main 2023, 24 Jan (Shift 2)]
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Let \(f\left(x\right)=x+\frac{a}{{\pi }^{2}-4}\sin x+\frac{b}{{\pi }^{2}-4}\cos x\), \(x\in R\) be a function which satisfies \(f(x)=x+\int_0^{\pi / 2} \sin (x+y) f(y) d y\). Then \((a+b)\) is equal to
[JEE Main 2023, 29 Jan (Shift 1)]
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If \([x]\) denotes the greatest integer less than or equal to \(x\), then the value of the integral \({\int }_{-\pi /2}^{\pi /2}[[x]-\sin x]dx\) is equal to:
[JEE Main 2021, 20 Jul (Shift 2)]
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Let \(f\) be a continuous function satisfying \(\int_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0\). Then \(f\left(\frac{\pi^2}{4}\right)\) is equal to:
[JEE Main 2023, 10 Apr (Shift 2)]
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The value of integral \({\int }_{0}^{\infty }\frac{6}{{e}^{3x}+6{e}^{2x}+11{e}^{x}+6}dx=\)
[JEE Main 2023, 13 Apr (Shift 1)]
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Which of the following statements is correct for the function \(g(\alpha )\) for \(\alpha \in R\) such that \(g(\alpha )={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\sin }^{\alpha }x}{{\cos }^{\alpha }x+{\sin }^{\alpha }x}dx\)
[JEE Main 2021, 17 Mar (Shift 1)]
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If \(f: R \rightarrow R\) be a continuous function satisfying \(\int_0^{\pi / 2} f(\sin 2 x) \cdot \sin x d x+\alpha \int_0^{\pi / 4} f(\cos 2 x) \cdot \cos x d x=0\), then \(\alpha\) is equal to :
[JEE Main 2023, 11 Apr (Shift 2)]
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If \( \theta_{1} \) and \( \theta_{2} \) be respectively the smallest and the largest values of \( \theta \) in \((0,2\pi )-\left\{\pi \right\}\) which satisfy the equation, \( 2 \cot ^{2} \theta-\frac{5}{\sin \theta}+4=0 \), then \( \int_{\theta_{1}}^{\theta_{2}} \cos ^{2} 3 \theta \mathrm{d} \theta \) is equal to:
[JEE Main 2020, 7 Jan (Shift 2)]
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The value of the definite integral \({\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{\left(1+{e}^{x\cos x}\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}\) is equal to
[JEE Main 2021, 27 Jul (Shift 1)]
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Let \(P(x)\) be a real polynomial of degree 3 which vanishes at \(x=-3\). Let \(P(x)\) have local minima at \(x=1\), local maxima at \(x=-1\) and \(\int_{-1}^1 P(x) d x=18\), then the sum of all the coefficients of the polynomial \(P(x)\) is equal to........
[JEE Main 2021, 18 Mar (Shift 2)]
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If \(I_n=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot ^n x d x\), then:
[JEE Main 2021, 25 Feb (Shift 2)]
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\(\lim _{n\to \infty }\frac{3}{n}\left\{4+{\left(2+\frac{1}{n}\right)}^{2}+{\left(2+\frac{2}{n}\right)}^{2}+\ldots +{\left(3-\frac{1}{n}\right)}^{2}\right\}\) is equal to
[JEE Main 2023, 30 Jan (Shift 2)]
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Let \(\mathrm{f}(\mathrm{x})\) be a differentiable function defined on \([0,2]\) such that \({f}^{'}(x)={f}^{'}(2-x)\) for all \(\mathrm{x}\in (0,2)\), \(f(0)=1\) and \(f(2)={e}^{2}\). Then the value of \({\int }_{0}^{2}f(x)dx\) is
[JEE Main 2021, 24 Feb (Shift 2)]
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\(\lim _{n\to \infty }\left[\frac{1}{n}+\frac{n}{(n+1{)}^{2}}+\frac{n}{(n+2{)}^{2}}+\ldots \ldots +\frac{n}{(2n-1{)}^{2}}\right]\) is equal to:
[JEE Main 2021, 25 Feb (Shift 2)]
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The value of the integral \({\int }_{-{\log }_{e}2}^{{\log }_{e}2}{e}^{x}\left({\log }_{e}\left({e}^{x}+\sqrt{1+{e}^{2x}}\right)\right)dx\) is equal to
[JEE Main 2023, 11 Apr (Shift 1)]
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The value of the integral \({\int }_{-1}^{1}{\log }_{e}(\sqrt{1-x}+\sqrt{1+x})dx\) is equal to:
[JEE Main 2021, 20 Jul (Shift 1)]
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If the real part of the complex number \((1-\cos \theta +2i\sin \theta {)}^{-1}\) is \(\frac{1}{5}\) for \(\theta \in (0,\pi )\), then the value of the integral \({\int }_{0}^{\theta }\sin xdx\) is equal to:
[JEE Main 2021, 20 Jul (Shift 2)]
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The value of the integral \(\int_{1 / 2}^2 \frac{\tan ^{-1} x}{x} d x\) is equal to
[JEE Main 2023, 29 Jan (Shift 2)]
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The minimum value of the function \(f\left(x\right)={\int }_{0}^{2}{e}^{|x-t|}dt\) is
[JEE Main 2023, 25 Jan (Shift 1)]
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For \(x>0\) ,If \( f(x)=\int_{1}^{x} \frac{\log _{e} t}{1+t} d t \) .Then \(f(e)+f\left(\frac{1}{e}\right)\) is equal to
[JEE Main 2021, 26 Feb (Shift 2)]
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Let \(f_1:(0, \infty) \rightarrow R\) and \(f_2:(0, \infty) \rightarrow R\) be defined by \(f_1(x)=\int_0^x \prod_{j=1}^{21}(t-j)^j d t, x>0\)
and \(f_2(x)=98(x-1)^{50}-600(x-1)^{49}+2450, x>0\), where for any positive integer \(n\) and real numbers \(a_1, a_2, \ldots a_n, \Pi_{i=1}^n a_i\) denotes the product of \(a_1, a_2, \ldots, a_n\). Let \(m_i\) and \(n_i\), respectively, denote the number of points of local minima and the number of points of local maxima of function \(f_i, i=1,2\), in the interval \((0, \infty)\)
The value of \(6 m_2+4 n_2+8 m_2 n_2\) is ________.
[JEE Advanced 2021]
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The integral \(16 \int_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}\) is equal to
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Let \( f \) be a differentiable function such that \( x^{2} f(x)-x=4 \int_{0}^{x} t f(t) d t, f(1)=\frac{2}{3} \). Then \( 18 f(3) \) is equal to
[JEE Main 2023, 10 Apr (Shift 1)]
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Which of the following statements is correct for the function \(g(\alpha)\) for \(\alpha \in R\) such that \(g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin ^\alpha x}{\cos ^\alpha x+\sin ^\alpha x} d x\)
[JEE Main 2021, 17 Mar (Shift 1)]
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\({\int }_{0}^{1}\frac{1}{\left(5+2x-2{x}^{2}\right)\left(1+{e}^{(2-4x)}\right)}dx=\frac{1}{\alpha }{\log }_{e}\left(\frac{\alpha +1}{\beta }\right).\) \(\alpha ,\beta ,>0\), then \({\alpha }^{4}-{\beta }^{4}\) is equal to:
[JEE Main 2023, 15 Apr (Shift 1)]
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If the integral \({\int }_{0}^{10}\frac{[\sin 2\pi x]}{{e}^{x-[x]}}dx=\alpha {e}^{-1}+\beta {e}^{-\frac{1}{2}}+\gamma\) where \(\alpha\), \(\beta, \gamma\) are integers and \([x]\) denotes the greatest integer less than or equal to \(x\), then the value of \(\alpha+\beta+\gamma\) is equal to:
[JEE Main 2021, 17 Mar (Shift 2)]
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The value of \(\alpha\) for which \(4\alpha {\int }_{-1}^{2}{e}^{-\alpha |x|}dx=5\) is
[JEE Main 2020, 7 Jan (Shift 2)]
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If \(\phi(x)=\frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0\), then \(\phi^{\prime}\left(\frac{\pi}{4}\right)\) is equal to:
[JEE Main 2023, 31 Jan (Shift 2)]
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Consider the integral \(I={\int }_{0}^{10}\frac{[x]{e}^{[x]}}{{e}^{x-1}}dx\), where \([x]\) denotes the greatest integer less than or equal to \(x\). Then the value of \(I\) is equal to:
[JEE Main 2021, 16 Mar (Shift 2)]
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If for all real triplets \((a, b, c), f(x)=a+b x+c x^2\); then \(\int_0^1 f(x) d x\) is equal to
[JEE Main 2020, 9 Jan (Shift 1)]
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The minimum value of the function \(f(x)={\int }_{0}^{2}{e}^{|x-t|}dt\) is
[JEE Main 2023, 25 Jan (Shift 1)]
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The integral \(\int_1^2 e^x \cdot x^x\left(2+\log _e x\right) d x\) equal :
[JEE Main 2020, 6 Sep (Shift 2)]
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The value of \({\int }_{-\pi /2}^{\pi /2}\frac{1}{1+{e}^{\sin x}}dx\) is
[JEE Main 2020, 5 Sep (Shift 1)]
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\(\int_6^{16} \frac{\log _e x^2}{\log _e x^2+\log _e\left(x^2-44 x+484\right)} d x\) is equal to.
[JEE Main 2021, 27 Aug (Shift 1)]
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If \( \theta_{1} \) and \( \theta_{2} \) be respectively the smallest and the largest values of \( \theta \) in \((0,2\pi )-\left\{\pi \right\}\) which satisfy the equation, \( 2 \cot ^{2} \theta-\frac{5}{\sin \theta}+4=0 \), then \(\int_{\theta_{1}}^{\theta_{2}} \cos ^{2} 3 \theta \mathrm{d} \theta \) is equal to:
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The value of the integral, \( \int_{1}^{3}\left[x^{2}-2 x-2\right] d x \), where \( [x] \) denotes the greatest integer less than or equal to \( x \) is:
[JEE Main 2021, 24 Feb (Shift 2)]
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The value of the integral \({\int }_{-1}^{1}\log \left(x+\sqrt{{x}^{2}+1}\right)dx\) is:
[JEE Main 2021, 25 Jul (Shift 2)]
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\(\int_{\pi / 3}^{\pi / 2} x \sin (\pi[x]-x) d x \text { is equal to : }\)
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The value of \(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^2}{n^2+4 r^2}\) is:
[JEE Main 2021, 26 Aug (Shift 1)]
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The value of \({\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left({\left(\frac{x+1}{x-1}\right)}^{2}+{\left(\frac{x-1}{x+1}\right)}^{2}-2\right)}^{1/2}dx\) is:
[JEE Main 2021, 26 Aug (Shift 1)]
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Let. \(5f(x)+4f\left(\frac{1}{x}\right)=\frac{1}{x}+3,x>0\) Then \(18{\int }_{1}^{2}f(x)dx\) is equal to:
[JEE Main 2023, 6 Apr (Shift 1)]
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If \(U_n=\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)^2 \ldots \ldots\left(1+\frac{n^2}{n^2}\right)^n\), then \(\lim _{n \rightarrow \infty}\left(U_n\right)^{\frac{-4}{n^2}}\) is equal to :
[JEE Main 2021, 27 Aug (Shift 1)]
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\( \int_0^{\infty} \frac{d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)} \text { is }\)
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\(\lim _{x \rightarrow 0} \frac{\int_0^x t \sin (10 t) d t}{x}\) is equal to
[JEE Main 2020, 8 Jan (Shift 2)]
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Let \(g(t)={\int }_{-\pi /2}^{\pi /2}\cos \left(\frac{\pi }{4}t+f(x)\right)dx\), where \(f(x)={\log }_{e}\left(x+\sqrt{{x}^{2}+1}\right),x\in R\). Then which one of the following is correct?
[JEE Main 2021, 20 Jul (Shift 2)]
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Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\pi^2, \forall x \in R\). Then \(\int_0^\pi f(x) \sin x d x\) is equal to
[JEE Main 2023, 6 Apr (Shift 2)]
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The value of the definite integral \({\int }_{\frac{\pi }{24}}^{\frac{5\pi }{24}}\frac{dx}{1+\sqrt[3]{\tan 2x}}\)
[JEE Main 2021, 25 Jul (Shift 1)]
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Let \( f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R} \) be a function which satisfies \( f(x)=x+\int_{0}^{\pi / 2} \sin (x+y) f(y) d y \). Then \( (a+b) \) is equal to
[JEE Main 2023, 29 Jan (Shift 1)]
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Let \(f:(0,1) \rightarrow R\) be the functions defined as \(f(x)=\sqrt{n}\) if \(x \in\left[\frac{1}{n+1}, \frac{1}{n}\right)\) where \(n \in N\). Let \(g :(0,1) \rightarrow R\) be a function such that \(\int_{x^2}^x \sqrt{\frac{1-t}{t}}d t [JEE Advanced 2023]
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Let \(f\) be a differentiable function defined on \(\left[0, \frac{\pi}{2}\right]\) such that \(f(x)>0\) and
\(f(x)+\int_0^x f(t) \sqrt{1-\left(\log _e f(t)\right)^2} d t=e, \forall x \in\left[0, \frac{\pi}{2}\right]\) then
\(\left(6 \log _e f\left(\frac{\pi}{6}\right)\right)^2\) is equal to
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\(\lim _{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}\) is equal to :
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Let \(\alpha>0\). If \(\int_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x=\frac{16+20 \sqrt{2}}{15}\), then \(\alpha\) is equal to :
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If \([\mathrm{x}]\) denotes the greatest integer \(\leq 1\), then the value of \(\frac{3(e-1)}{e}{\int }_{1}^{2}{x}^{2}{e}^{[x]+\left[{x}^{3}\right]}dx\) is:
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The integral \(\int_{\frac{\pi} { 6}}^{\frac{\pi} { 3}} \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x\) is equal to
[JEE Main 2019, 10 Apr (Shift 2)]
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The value of \({\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{(2+3\sin x)}{\sin x(1+\cos x)}dx\) is equal to
[JEE Main 2023, 31 Jan (Shift 1)]
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Let \(\alpha \in (0,1)\) and \(\beta ={\log }_{e}(1-\alpha )\).
Let \({P}_{n}(x)=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\ldots ..+\frac{{x}^{n}}{n},x\in (0,1)\).
Then the integral \({\int }_{0}^{\alpha }\frac{{t}^{50}}{1-t}dt\) is equal to
[JEE Main 2023, 31 Jan (Shift 1)]
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Let \(f(x)\) be a differentiable function defined on \([0,2]\) such that \({f}^{'}(x)={f}^{'}(2-x)\) for all \(x\in (0,2),f(0)=1\) and \(f\left(2\right)={e}^{2}\). Then the value of \({\int }_{0}^{2}f(x)dx\)is
[JEE Main 2021, 24 Feb (Shift 2)]
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Evaluate: \({\int }_{0}^{\pi }\frac{1}{5+4\cos x}dx\)
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Let \(g(x)=\int_0^x f(t) d t\), where \(f\) is continuous function in \([0,3]\) such that \(\frac{1}{3} \leq f(t) \leq 1\) for all \(t \in[0,1]\) and \(0 \leq f(t) \leq \frac{1}{2}\) for all \(t \in(1,3]\). The largest possible interval in which \(g (3)\) lies is:
[JEE Main 2021, 18 Mar (Shift 2)]
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Let \( f(x)=|x-2| \) and \( g(x)=f(f(x)), x \in[0,4] \). Then \( \int_{0}^{3}(\mathrm{~g}(\mathrm{x})-\mathrm{f}(\mathrm{x})) \mathrm{dx} \) is equal to:
[JEE Main 2020, 4 Sep (Shift 1)]
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Let \(f: R \rightarrow R\) be defined as \(f(x)=e^{-x} \sin x\). If \(F:[0,1] \rightarrow R\) is a differentiable function such that \(F(x)=\int_0^x f(t) d t\), then the value of \(\int_0^1\left(F^{\prime}(x)+f(x)\right) e^x d x\) lies in the interval.
[JEE Main 2021, 17 Mar (Shift 2)]
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The function \(f(x)\), that satisfies the condition \(f(x)=x+\int_0^{\pi / 2} \sin x \cdot \cos y f(y) d y\) is:
[JEE Main 2021, 1 Sep (Shift 2)]
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The value of the integral \({\int }_{-1}^{1}{\log }_{e}(\sqrt{1-x}+\sqrt{1+x})dx\) is equal to :
[JEE Main 2021, 20 Jul (Shift 1)]
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Evaluate: \(\ \int_0^\pi \frac{1}{5+4 \cos x} \mathbf{d x} \)
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\(\ \int_{-a}^4\left(x^8-x^4+x^2+1\right) d x=2 \int_0^4\left(x^8-x^4+x^2+1\right) d x \) then \(\ a= \)
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The integral \({\int }_{0}^{\frac{\pi }{2}}\frac{1}{3+2\sin x+\cos x}dx\) is equal to:
[JEE Main 2022, 29 Jul (Shift 1)]
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Which of the following statements is correct for the function \(g(\alpha )\) for \(\alpha \in R\) such that \(g(\alpha)=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin ^\alpha x}{\cos ^\alpha x+\sin ^\alpha x} d x\)
[JEE Main 2021, 17 Mar (Shift 1)]
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\(\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x\) is equal to
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\( \lim _{x \rightarrow 0^{+}} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) d t}{x^{3}} \) is equal to:
[JEE Main 2021, 24 Feb (Shift 1)]
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\({\int }_{6}^{16}\frac{{\log }_{e}{x}^{2}}{{\log }_{e}{x}^{2}+{\log }_{e}\left({x}^{2}-44x+484\right)}dx\) is equal to
[JEE Main 2021, 27 Aug (Shift 1)]
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Let \( f(x) \) and \( g(x) \) be two functions satisfying \( f\left(x^{2}\right)+g(4-x)=4 x^{3}, g(4-x)+g(x)=0 \), then the value of \( \int_{-4}^{4} f\left(x^{2}\right) \mathrm{d} x \) is
[JEE Main 2021, 18 Mar (Shift 1)]
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If \( f(a+b+1-x)=f(x) \), for all \( x \), where \( a \) and \( b \) are fixed positive real numbers, then \( \frac{1}{a+b} \int_{a}^{b} x(f(x)+f(x+1)) d x \) is equal to
[JEE Main 2020, 7 Jan (Shift 1)]
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The value of \({\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{{\cos }^{2}x}{1+{3}^{x}}dx\) is:
[JEE Main 2021, 26 Feb (Shift 1)]
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If the value of the integral \(\int_0^5 \frac{x+[x]}{e^{x-[x]}} d x=\alpha e^{-1}+\beta\), where \(\alpha, \beta \in R, 5 \alpha+6 \beta=0\), and \([x]\) denotes the greatest integer less than or equal to \(x\); then the value of \((\alpha+\beta)^2\) is equal to
[JEE Main 2021, 26 Aug (Shift 2)]
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Let \(a\) be a positive real numbers such that \(\int_0^a e^{x-[x]} d x=10 e-9\), where \([x]\) is the greatest integer less than or equal to \(x\). Then \(a\) is equal to:
[JEE Main 2021, 20 Jul (Shift 1)]
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The integral \( \int_{1}^{2} e^{x} \cdot x^{x}\left(2+\log _{e} x\right) d x \) equals :
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Let \(a\)be a positive real numbers such that \(\int_0^a e^{x-[x]} d x=10 e-9\) where \([x]\) is the greatest integer less than or equal to \(x\). Then \(a\) is equal to:
[JEE Main 2021, 20 Jul (Shift 1)]
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\(\lim _{x\to 0}\frac{48}{{x}^{4}}{\int }_{0}^{x}\frac{{t}^{3}}{{t}^{6}+1}dt\) is equal to _______ .
[JEE Main 2023, 30 Jan (Shift 1)]
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Let \(g(t)=\int_{\frac{-\pi} { 2}}^{\frac{\pi} { 2}} \cos \left(\frac{\pi}{4} t+f(x)\right) d x, \) where \(f(x)=\log _e\left(x+\sqrt{x^2+1}\right), x \in R \), Then which one of the following is correct?
[JEE Main 2021, 20 Jul (Shift 2)]
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Let. \(5f(x)+4f\left(\frac{1}{x}\right)=\frac{1}{x}+3,x>0\). Then \(18{\int }_{1}^{2}f\left(x\right)dx\) is equal to:
[JEE Main 2023, 6 Apr (Shift 1)]
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The value of \(\int_{-1}^1 x^2 e^{\left[x^3\right]} d x\), (where \([t]\) denotes the greatest integer \(\leq t\) ), is:
[JEE Main 2021, 25 Feb (Shift 1)]
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Let \(f:[0, \infty) \rightarrow[0, \infty)\) be defined as \(f(x)=\int_0^x[y] d y\)
where \([x]\) is the greatest integer less than or equal to \(x\). Which of the following is true?
[JEE Main 2021, 25 Jul (Shift 1)]
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The value of the integral \({\int }_{0}^{1}\frac{\sqrt{x}dx}{(1+x)(1+3x)(3+x)}\) is:
[JEE Main 2021, 27 Aug (Shift 2)]
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If \(f: R \rightarrow R\) is given by \(f(x)=x+1\), then the value of \(\lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots+f\left(\frac{5(n-1)}{n}\right)\right]\), is:
[JEE Main 2021, 20 Jul (Shift 2)]
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The value of the integral \(\int_{-\log _e 2}^{\log _e 2} e^x\left(\log _e\left(e^x+\sqrt{1+e^{2 x}}\right)\right) d x\) is equal to
[JEE Main 2023, 11 Apr (Shift 1)]
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Let a differentiable function \(f\) satisfy \(f(x)+\int_3^x \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3\). Then \(12 f(8)\) is equal to:
[JEE Main 2023, 31 Jan (Shift 1)]
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Let \(\alpha \in (0,1)\) and \(\beta ={\log }_{e}(1-\alpha )\).Let \({P}_{n}\left(x\right)=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\ldots ..+\frac{{x}^{n}}{n},x\in \left(0,1\right)\).Then the integral \({\int }_{0}^{\alpha }\frac{{t}^{50}}{1-t}dt\) is equal to
[JEE Main 2023, 31 Jan (Shift 1)]
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The value of \(\frac{e^{-\frac{\pi}{4}}+\int_0^{\frac{\pi}{4}} e^{-x} \tan ^{50} x d x}{\int_0^{\frac{\pi}{4}} e^{-x}\left(\tan ^{49} x+\tan ^{51} x\right) d x}\) is
[JEE Main 2023, 13 Apr (Shift 2)]
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If the value of the integral \({\int }_{0}^{1/2}\frac{{x}^{2}}{{\left(1-{x}^{2}\right)}^{3/2}}dx\) is \(\frac{k}{6}\), then \(k\) is equal to
[JEE Main 2020, 3 Sep (Shift 2)]
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The value of \(\frac{{e}^{-\frac{\pi }{4}}+{\int }_{0}^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx}{{\int }_{0}^{\frac{\pi }{4}}{e}^{-x}\left({\tan }^{49}x+{\tan }^{51}x\right)dx}\) is
[JEE Main 2023, 13 Apr (Shift 2)]
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If the real part of the complex number \((1-\cos \theta+2 i \sin \theta)^{-1}\) is \(\frac{1}{5}\) for \(\theta \in(0, \pi)\), then the value of the integral \(\int_0^\theta \sin x d x\) is equal to :
[JEE Main 2021, 20 Jul (Shift 2)]
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Let \( f \) be a continuous function satisfying \( \int_{0}^{t^{2}}\left(f(x)+x^{2}\right) \mathrm{dx}=\frac{4}{3} t^{3}, \forall t>0 \). Then \( \mathrm{f}\left(\frac{\pi^{2}}{4}\right) \) is equal to
[JEE Main 2023, 10 Apr (Shift 2)]
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Let \( f \) be a non-negative function defined on the interval \( [0,1] \). If \( \int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t, 0 \leq x \leq 1 \) and \( f(0)=0 \) then
[JEE Main 2021, 31 Aug (Shift 1)]
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The integral \( \int_{0}^{2}|| x-1|-x| d x \) is equal to
[JEE Main 2020, 2 Sep (Shift 1)]
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If for all real triplets \((a, b, c), f(x)=a+b x+c x^2\), then \(\int_0^1 f(x) d x\) is equal to
[JEE Main 2020, 9 Jan (Shift 1)]
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Let \(f:[0, \infty) \rightarrow[0, \infty)\) be defined as \(f(x)=\int_0^x[y] d y\), where \([x]\) is the greatest integer less than or equal to \(x\). Which of the following is true?
[JEE Main 2021, 25 Jul (Shift 1)]
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Let \(f(x)=x+\frac{a}{{\pi }^{2}-4}\sin x+\frac{b}{{\pi }^{2}-4}\cos x\), \(x\in R\) be a function which satisfies \(f(x)=x+{\int }_{0}^{\pi /2}\sin (x+y)f(y)dy\). Then \((a+b)\) is equal to
[JEE Main 2023, 29 Jan (Shift 1)]
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The value of \( \int_{0}^{2 \pi} \frac{x \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x \) is equal to:
[JEE Main 2020, 9 Jan (Shift 1)]
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\({\int }_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}\frac{48}{\sqrt{9-4{x}^{2}}}dx\) is equal to
[JEE Main 2023, 24 Jan (Shift 2)]
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The function \(f(x)\), that satisfies the condition \(f(x)=x+{\int }_{0}^{\pi /2}\sin x\cdot \cos yf(y)dy\) is:
[JEE Main 2021, 1 Sep (Shift 2)]
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Let \(a\) be a positive real numbers such that
\({\int }_{0}^{a}{e}^{x-[x]}dx=10e-9\)
where [x] is the greatest integer less than or equal to x. Then \(a\) is equal to :
[JEE Main 2021, 20 Jul (Shift 1)]
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If \( t \) denotes the greatest integer \( \leq t \), then the value of \( \frac{3(e-1)}{e} \int_{1}^{2} x^{2} e^{[x]+\left[x^{3}\right]} d x \) is equal to
[JEE Main 2023, 30 Jan (Shift 1)]
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\(\text{ The value of the definite integral }{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{\left(1+{e}^{x\cos x}\right)\left({\sin }^{4}x+{\cos }^{4}x\right)}\text{ is equal to : }\)
[JEE Main 2021, 27 Jul (Shift 1)]
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Let \(\alpha >0\). If \({\int }_{0}^{\alpha }\frac{x}{\sqrt{x+\alpha }-\sqrt{x}}dx=\frac{16+20\sqrt{2}}{15}\), then \(\alpha\) is equal to:
[JEE Main 2023, 31 Jan (Shift 2)]
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Let \(g_i:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow \mathbb{R}, i=1,2\), and \(f:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow \mathbb{R}\) be functions such that \(g_1(x)=1, g_2(x)=|4 x-\pi|\) and \(f(x)=\sin ^2 x\), for all \(x \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right]\) Define \( S_i=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f(x) \cdot g_i(x) d x, i=1,2 \) The value of \(\frac{48 S_2}{\pi^2}\) is
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The value of the integral, \(\int_1^3\left[x^2-2 x-2\right] d x\), where \([x]\) denotes the greatest integer less than or equal to \(x\) is
[JEE Main 2021, 24 Feb (Shift 2)]
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Let \( P(x)=x^{2}+b x+c \) be a quadratic polynomial with real coefficients such that \( \int_{0}^{1} P(x) d x=1 \) and \( P(x) \) leaves remainder 5 when it is divided by \( (x-2) \). Then the value of \( 9(b+c) \) is equal to:
[JEE Main 2021, 16 Mar (Shift 2)]
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If the integral \(\int_0^{10} \frac{[\sin 2 \pi x]}{e^{x-[x]}} d x=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma\) where \(\alpha\), \(\beta, \gamma\) are integers and \([x]\) denotes the greatest integer less than or equal to \(x\), then the value of \(\alpha+\beta+\gamma\) is equal to:
[JEE Main 2021, 17 Mar (Shift 2)]
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If \(f: R \rightarrow R\) be a continuous function satisfying \(\int_0^{\pi / 2} f(\sin 2 x) \cdot \sin x d x+\alpha \int_0^{\pi / 4} f(\cos 2 x) \cdot \cos x d x=0\), then \(\alpha\) is equal to
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The integral \({\int }_{\pi /6}^{\pi /3}{\tan }^{3}x\cdot {\sin }^{2}3x\left(2{\sec }^{2}x\cdot {\sin }^{2}3x+3\tan x.\sin 6x)dx\right.\) is equal to
[JEE Main 2020, 4 Sep (Shift 2)]
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\(\int _{0}^{∞}\frac{dx}{\left({x}^{2}+{a}^{2}\right)\left({x}^{2}+{b}^{2}\right)}\) is
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Let \(f(x)\) be a differentiable function defined on \([0,2]\) such that \({f}^{'}(x)={f}^{'}(2-x)\) for all \(x\in (0,2),f(0)=1\) and \(f\left(2\right)={e}^{2}\). Then the value of \( \int_0^2 f(x) d x\) is
[JEE Main 2021, 24 Feb (Shift 2)]
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If \(f:R\to R\) be a continuous function satisfying \({\int }_{0}^{\pi /2}f(\sin 2x)\cdot \sin xdx+\alpha {\int }_{0}^{\pi /4}f(\cos 2x)\cdot \cos xdx=0\),
then \(\alpha\) is equal to
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Let \(a\) differentiable function \(f\) satisfy
\[f(x)+\int_3^x \frac{f(t)}{t} d t=\sqrt{x+1}, x \geq 3 .\]
Then \(12 f(8)\) is equal to:
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