Definite Integration

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,

\(\displaystyle \int_0^2 \frac{1}{3^x+3}\,dx=\frac13\)

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 \(I=2-\sqrt{2}-\log_e(1+\sqrt{2})\).

$$\begin{gathered} F\left(x\right)=xf\left(x\right) \\ \Rightarrow F\left(x\right)=2f\left(2\right)=2 \end{gathered}$$

$$\begin{gathered} \text{Given that} {\int }_0^{2}x{F}^{\text{'}}\left(x\right)dx=6 \\ ={\left.xF\left(x\right)\right|}_0^2-{\int }_0^{2}F\left(x\right)dx=6 \left[\text{using integration by parts}\right] \\ =2 \mathrm{F}\left(2\right)-{\int }_0^{2}xf\left(x\right)\mathrm{dx}=6 \\ \Rightarrow {\int }_0^2 \mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}=-2 ....\left(1\right) \\ \mathrm{Also} \\ {\int }_0^2{x}^2{F}^{"}\left(x\right)dx={\left.x^2{F}^{\text{'}}\left(x\right)\right|}_0^2-2{\int }_0^{2}x{F}^{\text{'}}\left(x\right)dx=40 \\ =4{ \mathrm{F}}^{\text{'}}\left(2\right)-2\times 6=40 \\ {F}^{\text{'}}\left(2\right)=13 \\ \therefore {\mathrm{F}}^{\text{'}}\left(2\right)+{\int }_0^2 \mathrm{F}\left(\mathrm{x}\right)=13-2 \\ =11 \end{gathered}$$

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 \(\frac{1}{6}\)

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 \(2a+3b-4c=6-2+4=8\).

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 \(2a+3b-4c=6-2+4=8\).

$I_n={\int }_0^1{\left(1-x^k\right)}^n·1dx$

$I_n={\left[{\left(1-x^k\right)}^n·x\right]}_0^1-nk{\int }_0^1{\left(1-x^k\right)}^{n-1}\left(-x^{k-1}\right)·xdx$

$=-kn{\int }_0^1{\left(1-x^k\right)}^{n-1}\left(1-x^k-1\right)dx$

$I_n=-nk{\int }_0^1\left[{\left(1-x^k\right)}^n-{\left(1-x^k\right)}^{n-1}\right]dx$

$I_n=nk{I}_{n-1}-nk{I}_n$

$\frac{I_n}{I_{n-1}}=\frac{nk}{nk+1}$

$\frac{I_{21}}{I_{20}}=\frac{21k}{1+21k}$

$=\frac{147}{148}\Rightarrow k=7$

$I_n={\int }_0^1{\left(1-x^k\right)}^n·1dx$

$I_n={\left[{\left(1-x^k\right)}^n·x\right]}_0^1-nk{\int }_0^1{\left(1-x^k\right)}^{n-1}\left(-x^{k-1}\right)·xdx$

$=-kn{\int }_0^1{\left(1-x^k\right)}^{n-1}\left(1-x^k-1\right)dx$

$I_n=-nk{\int }_0^1\left[{\left(1-x^k\right)}^n-{\left(1-x^k\right)}^{n-1}\right]dx$

$I_n=nk{I}_{n-1}-nk{I}_n$

$\frac{I_n}{I_{n-1}}=\frac{nk}{nk+1}$

$\frac{I_{21}}{I_{20}}=\frac{21k}{1+21k}$

$=\frac{147}{148}\Rightarrow k=7$

Sol: $I = \int_0^\pi \frac{(x + 3)\sin x}{1 + 3\cos^2x} dx$

$$\begin{gathered} I={\int }_0^{\pi \mathrm{/}2}\frac{(x+3)\sin x}{1+3{\cos }^{2}x} \\ +\frac{(p-x+3)\sin (\pi -x)}{1+3{\cos }^2(\pi -x)}dx \end{gathered}$$

$$I = \int_0^{\pi/2} \frac{(\pi + 6)\sin x}{1 + 3\cos^2 x} dx$$

$$I - (\pi + 6) \int_0^{\pi/2} \frac{\sin x}{1 + 3\cos^2 x}$$

$$\begin{array}{l}I=(\pi +6){\int }_1^0\frac{-dt}{1+3t^2}(\because \cos x=t)\\ =\frac{\pi +6}{3}{\int }_0^1\frac{dt}{{\left(\frac{1}{\sqrt{3}}\right)}^2+t^2}\\ =\frac{\pi +6}{3}\cdot \sqrt{3}\cdot {\left({\tan }^{-1}\sqrt{3}t\right)}_0^1\\ =\frac{\pi +6}{\sqrt{3}}\cdot \frac{\pi }{3}\end{array}$$

$y=1-2x+e^{x}\int \limits _{0}^{x}e^{-t}f(t)dt$

${\tfrac {dy}{dx}}=-2+e^{-x}\cdot e^{x}f(x)+e^{x}\int \limits _{0}^{x}e^{-t}f(t)dt$

${\tfrac {dy}{dx}}=-2+y+y+2x-1$

${\tfrac {dy}{dx}}-2y=(2x-3)$

$I.F.=e^{\int -2dx}=e^{-2x}$

solution is

$ye^{-2x}=\int (2x-3)dx\cdot e^{-2x}$

$ye^{-2x}={\tfrac {-(2x-3)}{2}}e^{-2x}+\int e^{-2x}dx$

$ye^{-2x}={\tfrac {-(2x-3)}{2}}e^{-2x}-{\tfrac {1}{2}}e^{-2x}+c$

$\mathrm {f} (0)=1\Rightarrow \mathrm {c} =1-{\tfrac {3}{2}}+{\tfrac {1}{2}}=0$

$y=-{\tfrac {(2x-3)}{2}}-{\tfrac {1}{2}}$

$$\begin{gathered} y=-x+1 \\ \Rightarrow x+y=1 \end{gathered}$$

Hence area bounded by \(y=f(x)\) and coordinate axes is

area $=\frac{1}{2}(1)(1)=\frac{1}{2}$

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

\(L=\frac{-1\cdot \cos 0-1\cdot \sin 0}{6}=-\frac{1}{6}\).

$I={\pi }^2\underset{-1}{\overset{3/2}{\int }}\left|x\sin \pi x\right|dx$

$={\pi }^2\left\{\underset{-1}{\overset{1}{\int }}x\sin \pi xdx-\underset{1}{\overset{3/2}{\int }}x\sin \pi xdx\right\}$

$={\pi }^2\left\{2\underset{0}{\overset{1}{\int }}x\sin \pi xdx-\underset{-1}{\overset{3/2}{\int }}x\sin \pi xdx\right\}$

$\int x\sin \pi xdx-x\cdot \frac{1}{\pi }\cos \pi x+$$\int 1\cdot \frac{1}{\pi }\cos \pi xdx$

$={\pi }^2\left\{2\left(-\frac{x}{\pi }\cos \pi x+{\left.\frac{\sin \pi x}{{\pi }^2}\right)}_0^1\right.\right.$

$-\left(-\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^2}\right)$

$={\pi }^2\left\{\frac{2}{\pi }-\left(-\frac{1}{{\pi }^2}-\frac{1}{\pi }\right)\right\}$

$={\pi }^2\left\{{\left(-\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^2}\right)}_0^1\right.$

$-\left(-\frac{x}{\pi }\cos \pi x+\frac{\sin \pi x}{{\pi }^2}\right)$

$={\pi }^2\left\{\frac{2}{\pi }-\left(-\frac{1}{{\pi }^2}-\frac{1}{\pi }\right)\right\}$

$={\pi }^2\left\{\frac{3}{\pi }+\frac{1}{{\pi }^2}\right\}$

$=3\pi +1$

\(
\begin{aligned}
& \because f(x)=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} \\
& f^{\prime}(x)=\frac{2 x\left(x^4-8 x^2+15\right)}{e^{x^2}} \\
& =\frac{2 x\left(x^2-5\right)\left(x^2-3\right)}{e^{x^2}}
\end{aligned}
\)

The extremum value of \(f(x)\) are \(x=0, \pm \sqrt{5}, \pm \sqrt{3}\)
\(\therefore \quad\) Number of extremum points are 5.

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 \(t^3=u \Rightarrow 3 t^2 d t=d u\)

\(I=\left(\frac{2}{3}\right) \cdot \int_0^{2 \sqrt{2}}\left[\frac{d u}{\sqrt{u^2+1}}\right]\)

\(=\left(\frac{2}{3}\right) \cdot\left[\ln \left|u+\sqrt{u^2+1}\right|\right]_0^{2 \sqrt{2}}\)

\(=\left(\frac{2}{3}\right) \cdot[\ln (2 \sqrt{2}+3)-\ln 1]=\left(\frac{2}{3}\right) \cdot[\ln (3+2 \sqrt{2})]\)

$f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x$$-3{\tan }^{2}x$

$f\left(x\right)=7{\tan }^{6}x\left({\tan }^{2}x+1\right)$$-3{\tan }^{2}x\left({\tan }^{2}x+1\right)$

$f\left(x\right)=\left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}x$

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 $\tan x=t\Rightarrow dx={\sec }^{2}t dt$

$$\begin{gathered} I_1={\int }_0^1\left(7t^6-3t^2\right)dt={\left[t^7-t^3\right]}_0^1=0 \\ {I}_2={\int }_0^{\pi /4}x\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left({\sec }^{2}x\right)dx \\ ={\left[x\left({\tan }^{7}x-{\tan }^{3}x\right)\right]}_0^{\pi /4}-{\int }_0^{\pi /4}\left({\tan }^{7}x-{\tan }^{3}x\right)dx \end{gathered}$$

$=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 \(\tan x=t \Rightarrow d x=\sec ^2 t d t\)

$$\begin{gathered} =-{\int }_0^1\left(t^5-t^3\right)dt=-{\left[\frac{t^6}{6}-\frac{t^4}{4}\right]}_0^1=\frac{1}{12} \\ 7{\mathrm{I}}_1+12{\mathrm{I}}_2=1 \end{gathered}$$

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 \(I=\sqrt{2}-\sqrt{5}+\log_e\left(\frac{9+4\sqrt{5}}{1+\sqrt{2}}\right)\).

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

$I_2={\int }_0^{\pi /2}\frac{\left({\displaystyle \frac{\pi }{2}-x}\right)\sin x\cos x}{{\sin }^{4}x+{\cos }^{4}x}dx$

and adding both, we get

$I_2=\frac{\pi }{4}{\int }_0^{\pi /2}\frac{\tan x{\sec }^{2}xdx}{{\tan }^{4}x+1}$

$\text{ put }{\tan }^2\mathrm{x}=\mathrm{t}\Rightarrow \mathrm{tanx} {\sec }^2\mathrm{x} \mathrm{dx}=\frac{\mathrm{dt}}{2}$

$I_2=\frac{\mathrm{\pi }}{8}{\int }_0^{\infty }\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}$

$$\begin{gathered} =\frac{\mathrm{\pi }}{8}{\left[{\tan }^{-1}\mathrm{t}\right]}_0^{\infty } \\ =\frac{\mathrm{\pi }}{8}·\frac{\mathrm{\pi }}{2}=\frac{{\mathrm{\pi }}^2}{16} \end{gathered}$$

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 \(\frac{\sqrt2\pi^2}{32}\)

\( I=\int_0^1 \cot ^{-1}\left(1+x+x^2\right) d x\)

\(=\int_0^1 \tan ^{-1}\left(\frac{1}{1+x+x^2}\right) d x\)

\(=\int_0^1 \tan ^{-1}\left(\frac{x+1-x}{1+x(x+1)}\right) d x\)

\(I=\int_0^1 \tan ^{-1}(x+1) d x-\int_0^1 \tan ^{-1} x d x\)

\(=\int_1^2 \tan ^{-1} x d x-\int_0^1 \tan ^{-1} x d x\)

\(\because \int \tan ^{-1} x d x=x \tan ^{-1} x-\frac{1}{2} \ln \left(1+x^2\right)\)

\(\Rightarrow I=2 \tan ^{-1} 2-\frac{1}{2} \ln \left(\frac{5}{4}\right)-\frac{\pi}{2}\)

$$f(x)=\log_2\log_4\log_6(3+4x-x^2)$$

$$\log_6(3+4x-x^2)>0\Rightarrow 3+4x-x^2>1$$

$$-x^2+4x+2>0\Rightarrow x^2-4x-2<0$$

Roots: $x=2\pm\sqrt6$.

So, $$2-\sqrt6<x<2+\sqrt6.$$

${\log }_4\left({\log }_6\right(3+4x-x^2\left)\right)>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 $1<x<3$.

$$\log_2(\log_4\log_6(\cdot))\ \text{requires}\ \log_4\log_6(\cdot)>0,$$

which is the same as step 2. So

$$\text{domain}=(a,b)=(1,3).$$

Thus $b-a=2$ and we need

$$\int_{0}^{2} \lfloor x^2\rfloor\,dx.$$

On $[0,2]$, $x^2$ runs from 0 to 4; break at where $x^2=k$:

• $0\le x<1$: $\lfloor x^2\rfloor=0$
• $1\le x<\sqrt2$: $\lfloor x^2\rfloor=1$
• $\sqrt2\le x<\sqrt3$: $\lfloor x^2\rfloor=2$
• $\sqrt3\le x\le2$: $\lfloor x^2\rfloor=3$

So

${\int }_0^2\lfloor x^2\rfloor dx={\int }_0^{1}0dx+{\int }_1^{\sqrt{2}}1dx$$+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$

$$=(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3).$$

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

$$p+q+r=5+2+3=10.$$

Roots are \(1 \pm \mathrm{i} \sqrt{2} \) and \(1\)

Sum of roots:

\(-\mathrm{a}=3 \Rightarrow \mathrm{a}=-3\)

Product of roots:

\(-\mathrm{c}=1(1+\sqrt{2} \mathrm{i})(1-\mathrm{i} \sqrt{2})\)

\(\Rightarrow \mathrm{c}=-3\)

\(I=\int_{-1}^1\left(x^3-3 x^2+b x-3\right) d x\)

\(=2 \int_0^1\left(-3 x^2-3\right) d x\)

\(=2\left(-x^3-3 x\right)_0^1=-8\)

\(L=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)\)

Applying L-Hospital's Rule

$\Rightarrow L=\underset{x\to \frac{\pi }{2}}{lim} \frac{-\cos \left(x\right)3x^2}{2\left(x-\frac{\pi }{2}\right)}=\underset{x\to \frac{\pi }{2}}{lim} \frac{\sin \left(x-\frac{\pi }{2}\right)3x^2}{2\left(x-\frac{\pi }{2}\right)}$

$\Rightarrow L=\frac{3{\pi }^2}{8}$

\(L=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^{2}} \int_{x^{3}}^{\left(\frac{\pi}{2}\right)^{3}} \cos \left(t^{\frac{1}{3}}\right) d t\right)\)

Applying L-Hospital's Rule

$\Rightarrow L=\underset{x\to \frac{\pi }{2}}{lim} \frac{-\cos \left(x\right)3x^2}{2\left(x-\frac{\pi }{2}\right)}=\underset{x\to \frac{\pi }{2}}{lim} \frac{\sin \left(x-\frac{\pi }{2}\right)3x^2}{2\left(x-\frac{\pi }{2}\right)}$

$\Rightarrow L=\frac{3{\pi }^2}{8}$

$$\begin{gathered} \text{Let} I=80{\int }_0^{\frac{\pi }{4}}\left(\frac{\sin \theta +\cos \theta }{9+16(2\sin \theta ·\cos \theta )}\right)d\theta \\ =80{\int }_0^{\frac{\pi }{4}}\frac{\sin \theta +\cos \theta }{9+16-16(\sin \theta -\cos \theta {)}^2}d\theta \\ \text{Put} \sin \theta -\cos \theta =t \\ \left(\cos \theta +\sin \theta \right)\mathrm{d}\theta =dt \\ I=80{\int }_{-1}^0\frac{\mathrm{dt}}{25-16t^2} \\ =\frac{80}{16}{\int }_{-1}^0\frac{\mathrm{dt}}{{\left(\frac{5}{4}\right)}^2-{\mathrm{t}}^2}{\left.=\frac{5}{2\left(\frac{5}{4}\right)}\ln \left(\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right)\right]}_{-1}^0 \\ =4\ln 3 \end{gathered}$$

Sol. \(I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)..(i)
By property - 5

\(\int_0^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)..(ii)

Add eq. (i) and (ii)

\(
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}} \mathrm{dx} \\
& 2 \mathrm{I}=\frac{\pi}{2} \\
& \mathrm{I}=\frac{\pi}{4}
\end{aligned}
\)

Let \(J=\int_0^{2 I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\)

\(
J=\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x ..................(i)
\)

By property - 5

\(
J=\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x ............(ii)
\)

\(\begin{aligned}
&\text { Adding, }\\
&2 J=\int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \cos x \sin x d x}{\left(\sin ^4 x+\cos ^4 x\right)}
\end{aligned}\)

\(\begin{aligned}
& \left(\sin ^4 x+\cos ^4 x\right)=\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x \\
& =\left(1-\frac{\sin ^2 2 x}{2}\right)=\frac{2-\sin ^2 2x}{2} \\
& =\frac{1-\cos ^2 2 x}{2} \\
& 2 J=\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^2 2 x} \\
& \Rightarrow 2 J=\frac{\pi}{4} \int_{-1}^1 \frac{d t}{\left(1+t^2\right)^2}, t=\cos 2 x \\
& \Rightarrow J=\frac{\pi}{8} \cdot\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)=\frac{\pi^2}{16}
\end{aligned}\)

$\mathrm{f}\left(\mathrm{a}\right)=\frac{3^{\mathrm{a}}+\frac{1}{3^{\mathrm{a}}}+2\left(2^{\mathrm{a}}+\frac{1}{2^{\mathrm{a}}}\right)}{2}$

\( \mathrm{f}(\mathrm{a})_{\min }=3 \) at \( \mathrm{a} =0 \)

\( \alpha=3 \)

\( \mathrm{I}=\int_{\ln 2}^{\ln 3} \frac{\mathrm{dx}}{\mathrm{e}^{2 \mathrm{x}}-\mathrm{e}^{-2 \mathrm{x}}} \)

Let \( \mathrm{e}^{2 \mathrm{x}}=\mathrm{t}\)

\( \Rightarrow 2 \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}=\mathrm{dt} \)

\( \mathrm{I}=\frac{1}{2} \int_4^9 \frac{\mathrm{dt}}{\mathrm{t}^2-1}=\frac{1}{4}\left(\mathrm{ln} \frac{4}{5}-\mathrm{ln} \frac{3}{5}\right)\)

\(=\frac{1}{4}\left(\ln \frac{4}{3}\right)\)

$\Rightarrow I=\underset{-2}{\overset{2}{\int }}\frac{9x^2}{1+5^x}dx=\mathrm{5.....}\left(1\right)$

Taking $I=\underset{-2}{\overset{2}{9\int }}\frac{5^x{x}^2}{1+5^x}dx\mathrm{.....}\left(2\right)$

Add (1) + (2)

$\Rightarrow 2I=\underset{-2}{\overset{2}{\int }}9x^{2}dx$

${\left.=3x^3\right|}_{-2}^2=48$

I = 24

Left-hand side (LHS): By FTC, the derivative of $10 \int_{1}^{x} f(t) \, dt$ with respect to $x$ is $10 f(x)$.

Right-hand side (RHS): Differentiate $5x f(x) - x^5 - 9$ using the product rule and power rule:

$$\frac{d}{dx} \left( 5x f(x) - x^5 - 9 \right) = 5 f(x) + 5x f'(x) - 5x^4.$$

Equating the derivatives:

$$10 f(x) = 5 f(x) + 5x f'(x) - 5x^4.$$

$$f(x) = x f'(x) - x^4.$$

Rearrange to standard linear form:

$$x f'(x) - f(x) = x^4 \implies f'(x) - \frac{1}{x} f(x) = x^3.$$

The integrating factor is $\mu(x) = e^{\int -\frac{1}{x} dx} = \frac{1}{x}$. Multiply both sides by $\mu(x)$:

$$\frac{1}{x} f'(x) - \frac{1}{x^2} f(x) = x^2.$$

The left-hand side is the derivative of $\frac{f(x)}{x}$:

$$\frac{d}{dx} \left( \frac{f(x)}{x} \right) = x^2.$$

Integrate both sides:

$$\frac{f(x)}{x} = \int x^2 dx + C = \frac{x^3}{3} + C.$$

Multiply by $x$:

$$f(x) = \frac{x^4}{3} + C x.$$

The integral $\int_{1}^{1} f(t) dt = 0$, so:

$$0 = 5 f(1) - 1 - 9 \implies 5 f(1) = 10 \implies f(1) = 2.$$

Substitute $x = 1$ into $f(x)$:

$$f(1) = \frac{1^4}{3} + C \cdot 1 = \frac{1}{3} + C = 2 \implies C = \frac{5}{3}.$$

$$f(x) = \frac{x^4}{3} + \frac{5}{3} x = \frac{1}{3} (x^4 + 5x).$$

$$f(3)=\frac{1}{3}(3^4+5\cdot 3)=\frac{1}{3}(81+15)=\frac{96}{3}=32.$$

Let \(I_1=\int_{-\frac{1}{2}}^1 2xf(2x(1-2x))dx\).

Put \(x=\frac{1}{2}-t\), then \(2x(1-2x)\)

Since \(2\left(\frac{1}{2}-t\right)\left(1-2\left(\frac{1}{2}-t\right)\right)=2t(1-2t)\).

So \(I_1=\int_{-\frac{1}{2}}^1 (1-2x)f(2x(1-2x))dx\).

Adding this with the original form of \(I_1\),

we get \(2I_1=\int_{-\frac{1}{2}}^1 f(2x(1-2x))dx\).

Hence \(I_1=\frac{1}{2}\int_{-\frac{1}{2}}^1 f(2x(1-2x))dx\).

Now put \(u=2x\), then \(dx=\frac{du}{2}\).

When \(x=-\frac{1}{2}\), \(u=-1\), and when \(x=1\), \(u=2\).

Also \(2x(1-2x)=u(1-u)\).

Therefore \( \int_{-\frac{1}{2}}^1 f(2x(1-2x))dx=\frac{1}{2}\int_{-1}^2 f(u(1-u))du=\frac{I_2}{2} \).

So \(I_1=\frac{1}{2}\cdot\frac{I_2}{2}=\frac{I_2}{4}\).

Therefore \( \frac{I_2}{I_1}=4 \).

$$\begin{gathered} I={\int }_{{\mathrm{e}}^2}^{{\mathrm{e}}^4}\frac{1}{\mathrm{x}}\left(\frac{{\mathrm{e}}^{{\left({\left({\log }_{\mathrm{e}}\mathrm{x}\right)}^2+1\right)}^{-1}}}{{\mathrm{e}}^{{\left({\left({\log }_{e}x\right)}^2+1\right)}^{-1}}+{\mathrm{e}}^{{\left({\left(6-{\log }_{e}x\right)}^2+1\right)}^{-1}}}\right)dx \\ \text{let} {\log }_{e}x=t \\ \frac{dx}{x}=dt \\ \text{At} x=e^2, t = 2 \\ \text{At} x=e^4, t = 4 \\ \text{Now, } \\ I={\int }_2^4\frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}}+e^{\frac{1}{1+(6-t{)}^2}}}dt ........\left(i\right) \\ I={\int }_2^4\frac{e^{\frac{1}{1+(6-t{)}^2}}}{e^{\frac{1}{1+(6-t{)}^2}}+e^{\frac{1}{1+t^2}}}dt .........\left(ii\right) \\ \text{Add eqn. }\left(i\right) \& \left(ii\right), we get \\ 2I ={\int }_2^4\mathrm{dt}={\left(\mathrm{t}\right)}_2^4=4-2=2 \\ I=1 \end{gathered}$$

$$\begin{gathered} I={\int }_0^{\pi /2}\left(\frac{96\left(x^2+\cos x\right)}{1+e^x}+\frac{96\left((-x{)}^2+\cos (-x)\right)}{1+e^{-x}}\right)dx \\ ={\int }_0^{\pi /2}96\left(x^2+\cos x\right)\left[\frac{1}{1+e^x}+\frac{e^x}{1+e^x}\right]dx \\ ={\int }_0^{\pi /2}96\left(x^2+\cos x\right)\left[\frac{1+e^x}{1+e^x}\right]dx \\ =96{\left[\frac{x^3}{3}+\sin x\right]}_0^{\pi /2}=96\left[\frac{{\pi }^3}{24}+1\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow 4{\mathrm{\pi }}^3+96 ={\mathrm{\alpha \pi }}^3+\mathrm{\beta } \\ \mathrm{\alpha }=4, \mathrm{\beta }=96 \\ \mathrm{then} \mathrm{\alpha }+\mathrm{\beta } = 4+96=100 \end{gathered}$$

$$\begin{gathered} \text{Given that }\mathrm{g}\left(x^3\right)=x^6+x^7 \\ g\left(x\right)=x^2+x^{\frac{7}{3}} \\ {g}^{\text{'}}\left(x\right)=2x+\frac{7}{3}{x}^{\frac{4}{3}} \\ g\left(x\right)={\int }_0^{x}tf\left(t\right)dt \\ {g}^{\text{'}}\left(x\right)=xf\left(x\right) \\ f\left(x\right)=\frac{g^{\text{'}}\left(x\right)}{x} \\ \mathrm{f}\left(\mathrm{x}\right)=2+\frac{7}{3}{\mathrm{x}}^{\frac{1}{3}} \\ \mathrm{f}\left({\mathrm{r}}^3\right)=2+\frac{7\mathrm{r}}{3} \\ \sum _{\mathrm{r}=1}^{15}\left(2+\frac{7}{3}\mathrm{r}\right)=310 \end{gathered}$$

Given, \(\mathrm{I}(\mathrm{m}, \mathrm{n})=\int_0^1 \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{dx}, \mathrm{m}, \mathrm{n}>0 \ldots(i)\)
let, \(\sin ^2 t=x \Rightarrow 2 \sin t \cdot \cos t d t=d x\)
at \(x=0, t=0\)
at \(x=1, t=\pi / 2\)
\(\mathrm{I}(\mathrm{m}, \mathrm{n})=2 \int_0^{\pi / 2}(\sin t)^{2 m-1}(\cos t)^{2 \mathrm{n}-1} \mathrm{dt}\)
Now, put \(\mathrm{m}=9, \mathrm{n}=14\)
\(\mathrm{I}(9,14)=2 \int_0^{\pi / 2}(\sin \mathrm{t})^{17}(\cos \mathrm{t})^{27} \mathrm{dt} \ldots (ii)\)
put \(\mathrm{m}=10, \mathrm{n}=13\)
\(\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \mathrm{t})^{19}(\cos \mathrm{t})^{25} \mathrm{dt} \ldots (iii)\)
Add equation (ii) & (iii), we get

\( \mathrm{I}(9,14)+\mathrm{I}(10,13)= 2 \int_0^{\pi / 2}(\sin \mathrm{t})^{17}(\cos \mathrm{t})^{27} \mathrm{dt}+2 \int_0^{\pi / 2}(\sin \mathrm{t})^{19}(\cos \mathrm{t})^{25} \mathrm{dt} \)
\( \Rightarrow \mathrm{I}(9,14)+\mathrm{I}(10,13)=2 \int_0^{\pi / 2}(\sin \mathrm{t})^{17}(\cos \mathrm{t})^{25} \mathrm{dt} \)
\( =\mathrm{I}(9,13)\)

Sol. \(I=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)..(i)
By property - 5

\(\int_0^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x\)..(ii)

Add eq. (i) and (ii)

\(
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}} \mathrm{dx} \\
& 2 \mathrm{I}=\frac{\pi}{2} \\
& \mathrm{I}=\frac{\pi}{4}
\end{aligned}
\)

Let \(J=\int_0^{2 I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\)

\(
J=\int_0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x ..................(i)
\)

By property - 5

\(
J=\int_0^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x ............(ii)
\)

\(\begin{aligned}
&\text { Adding, }\\
&2 J=\int_0^{\frac{\pi}{2}} \frac{\frac{\pi}{2} \cos x \sin x d x}{\left(\sin ^4 x+\cos ^4 x\right)}
\end{aligned}\)

\(\begin{aligned}
& \left(\sin ^4 x+\cos ^4 x\right)=\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x \\
& =\left(1-\frac{\sin ^2 2 x}{2}\right)=\frac{2-\sin ^2 2x}{2} \\
& =\frac{1-\cos ^2 2 x}{2} \\
& 2 J=\frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^2 2 x} \\
& \Rightarrow 2 J=\frac{\pi}{4} \int_{-1}^1 \frac{d t}{\left(1+t^2\right)^2}, t=\cos 2 x \\
& \Rightarrow J=\frac{\pi}{8} \cdot\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)=\frac{\pi^2}{16}
\end{aligned}\)

For \(f(x)=\log_2\log_4\log_6(3+4x-x^2)\),

\( \log_4\log_6(3+4x-x^2)>0 \).

Since base \(4>1\), this gives \( \log_6(3+4x-x^2)>1 \).

Since base \(6>1\), this gives \(3+4x-x^2>6\).

So \(-x^2+4x-3>0\), which gives \(x^2-4x+3<0\).

Thus \((x-1)(x-3)<0\), so the domain is \((1,3)\).

Therefore \(a=1\), \(b=3\), and \(b-a=2\).

Now \( \int_0^{b-a}[x^2]dx=\int_0^2[x^2]dx \).

For \(0\le x<1\), \([x^2]=0\).

For \(1\le x<\sqrt{2}\), \([x^2]=1\).

For \(\sqrt{2}\le x<\sqrt{3}\), \([x^2]=2\).

For \(\sqrt{3}\le x\le 2\), \([x^2]=3\).

Therefore \( \int_0^2[x^2]dx=0(1-0)+1(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3}) \).

So \( \int_0^2[x^2]dx=\sqrt{2}-1+2\sqrt{3}-2\sqrt{2}+6-3\sqrt{3} \).

Hence \( \int_0^2[x^2]dx=5-\sqrt{2}-\sqrt{3} \).

Comparing with \(p-\sqrt{q}-\sqrt{r}\), we get \(p=5\), \(q=2\), and \(r=3\).

Therefore \(p+q+r=5+2+3=10\).

Let \(I=\int_0^{\frac{\pi}{4}} \frac{136\sin x}{3\sin x+5\cos x}dx\).

We write \(136\sin x=A(3\sin x+5\cos x)+B(3\cos x-5\sin x)\).

Comparing coefficients of \(\sin x\) and \(\cos x\),

we get \(3A-5B=136\) and \(5A+3B=0\).

On solving \(B=-20\) and \(A=12\).

So \(136\sin x=12(3\sin x+5\cos x)-20(3\cos x-5\sin x)\).

Hence \( \frac{136\sin x}{3\sin x+5\cos x}=12-20 \left(\frac{3\cos x-5\sin x}{3\sin x+5\cos x}\right) \).

Therefore \(I=\int_0^{\frac{\pi}{4}}\left(12-20\left(\frac{3\cos x-5\sin x}{3\sin x+5\cos x}\right)\right)dx\).

So \(I=\left[12x-20\log_e(3\sin x+5\cos x)\right]_0^{\frac{\pi}{4}}\).

Now at \(x=\frac{\pi}{4}\), \(3\sin x+5\cos x=\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}=4\sqrt{2}\).

At \(x=0\), \(3\sin x+5\cos x=5\).

Thus \(I=3\pi-20\log_e\left(\frac{4\sqrt{2}}{5}\right)\).

So \(I=3\pi-20\log_e(4\sqrt{2})+20\log_e 5\).

Since \( \log_e(4\sqrt{2})=\log_e(2^2\cdot 2^{\frac{1}{2}})=\frac{5}{2}\log_e 2 \),

we get \(I=3\pi-50\log_e 2+20\log_e 5\).

Let \(I=\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y}dy\).

Now \(I=\int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y}dy+\int_{-\pi}^{\pi} \frac{2y\sin y}{1+\cos^2 y}dy\).

Since \( \frac{2y}{1+\cos^2 y} \) is an odd function,

\( \int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y}dy=0 \).

Also \( \frac{2y\sin y}{1+\cos^2 y} \) is an even function,

so \(I=2\int_0^\pi \frac{2y\sin y}{1+\cos^2 y}dy=4\int_0^\pi \frac{y\sin y}{1+\cos^2 y}dy\).

Let \(J=\int_0^\pi \frac{y\sin y}{1+\cos^2 y}dy\).

\(J=\left[-y\tan^{-1}(\cos y)\right]_0^\pi+\int_0^\pi \tan^{-1}(\cos y)dy\).

Now \( \left[-y\tan^{-1}(\cos y)\right]_0^\pi=-\pi\tan^{-1}(-1)-0=\frac{\pi^2}{4} \).

Also \( \tan^{-1}(\cos(\pi-y))=\tan^{-1}(-\cos y)=-\tan^{-1}(\cos y) \),

so \( \int_0^\pi \tan^{-1}(\cos y)dy=0 \).

Therefore \(J=\frac{\pi^2}{4}\).

Hence \(I=4J=4\cdot \frac{\pi^2}{4}=\pi^2\).

$$\begin{gathered} \text{Given; }{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96x^2{\cos }^{2}x}{\left(1+e^x\right)}dx=\pi \left(\alpha {\pi }^2+\beta \right),\alpha ,\beta \in Z, \\ \text{let}, I ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96x^2{\cos }^{2}x}{\left(1+e^x\right)}dx ...\left(i\right) \end{gathered}$$

On applying king's property, we get

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{96x^2{\cos }^{2}x}{1+e^{-x}} dx ...\left(ii\right)$

adding (i) and (ii), we get

$$\begin{gathered} 2I = 2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}96x^2{\cos }^{2}xdx \\ 2I = 2{\int }_0^{\frac{\pi }{2}}96x^2{\cos }^{2}xdx \\ I=48{\int }_0^{\frac{\pi }{2}}{x}^2\left(1+\cos 2x\right)dx \\ I=48\left[{\left(\frac{x^3}{3}\right)}_0^{\pi /2}+{\int }_0^{\frac{\pi }{2}}{x}^2\cos 2xdx\right] \\ I=48\left[{\left(\frac{x^3}{3}\right)}_0^{\pi /2}+{\left(\frac{x^2-\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}\right)}_0^{\pi /2}\right] \\ I = \pi \left(2{\pi }^2-12\right) \\ \text{On comapring with} \pi \left(\alpha {\pi }^2+\beta \right), \text{we get} \\ \alpha =2, \beta =-12 \\ \text{Now}, {\left(\alpha +\beta \right)}^2=100 \end{gathered}$$