Indefinite Integration
9 Board Maths previous year questions on Indefinite Integration — options free on every question; 1 include the answer & explanation free, the rest unlock with PYQ Pass.
\(\int \frac{{\sin }^{2}x−{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx\text{ is equal to : }\)
\(\tan \left(x\right)+cotx+c\)
Given integral: \(I=\int \frac{(\sin ²x-\cos ²x)}{(\sin ²x\cos ²x)}dx\)
Using the identity: \(\sin ²x-\cos ²x=-\cos 2x\)
The integral becomes: \(I=\int \frac{(-\cos 2x)}{(\sin ²x\cos ²x)}dx\)
Using the identity:\(\sin ²x\cos ²x=(\frac{1}{4})\sin ²2x\)
The integral transforms into: \(I=-4\int \frac{\cos 2x}{\sin ²2x}dx\)
Substituting t=sin2x, so that \(dt=2\cos 2xdx\)
Thus, \(dx=\frac{dt}{(2\cos 2x)}\)
Substituting in the integral:
\(I=-4\int (\frac{\cos 2x}{t²})\times (\frac{dt}{(2\cos 2x)})\\ =-4\int (\frac{dt}{(2t²)})\\ =-2\int t⁻²dt\\ =\frac{2}{t}+c\)
Using the identity we get: \(\frac{1}{\sin x\cos x}=\tan x+cotx\)+ c
\(\int \frac{1}{x{\left(\log x\right)}^{2}}\mathrm{d}x\) is equal to :
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\(\int \frac{{e}^{x}(1+x)}{{\cos }^{2}\left(x⋅{e}^{x}\right)}dx\text{ is equal to : }\)
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\(\int \frac{{\sin }^{2}x−{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx\text{ is equal to : }\)
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\(\int \frac{1}{x{\left(\log x\right)}^{2}}\mathrm{d}x\) is equal to :
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\(\int \frac{{e}^{x}(1+x)}{{\cos }^{2}\left(x⋅{e}^{x}\right)}dx\text{ is equal to : }\)
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\(\begin{aligned}& f(x)=7(\tan x)^8+7(\tan x)^6-3(\tan x)^4-3\left(\tan ^2 x\right) \\& I_1=\int f(x) d x, I_2= \int x f(x) d x \\& 7 I_1+12 I_2=\end{aligned}\)
[Jee Main 2025]
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The value of \(\int {\cos }^{2}xdxis-\)
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If \(\ \int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+C \) , then value of \(\ (A, B) \) is :
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