NCERT Solutions · Class 12 Mathematics Integrals

Let \(\displaystyle I=\displaystyle\int e^{2x}\sin x\,dx\). Apply parts twice; the original integral reappears and is then solved for algebraically. First, with \(\displaystyle \sin x\) as the first function and \(\displaystyle e^{2x}\) as the second: \[I = \frac{e^{2x}}{2}\sin x-\frac12\int e^{2x}\cos x\,dx\] Apply parts to the new integral, again with the trigonometric factor first: \[\int e^{2x}\cos x\,dx = \frac{e^{2x}}{2}\cos x+\frac12\int e^{2x}\sin x\,dx = \frac{e^{2x}}{2}\cos x+\frac{I}{2}\] Substituting back: \[I = \frac{e^{2x}}{2}\sin x-\frac{e^{2x}}{4}\cos x-\frac{I}{4}\quad\Rightarrow\quad \frac{5I}{4} = \frac{e^{2x}}{4}\left(2\sin x-\cos x\right)\] \[I = \frac{e^{2x}}{5}\left(2\sin x-\cos x\right)+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{e^{2x}}{5}\left(2\sin x-\cos x\right)+\mathrm{C}\).

First simplify the inverse function. Put \(\displaystyle x=\tan\theta\), i.e. \(\displaystyle \theta=\tan^{-1}x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\). Then \[\frac{2x}{1+x^{2}} = \frac{2\tan\theta}{1+\tan^{2}\theta} = \sin 2\theta\] Now \(\displaystyle \sin^{-1}(\sin 2\theta)=2\theta\) only when \(\displaystyle 2\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), i.e. \(\displaystyle \theta\in\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]\), i.e. \(\displaystyle -1\le x\le 1\). This range check is the step that decides the answer; taking it as \(\displaystyle 2\tan^{-1}x\) for all \(\displaystyle x\) is wrong. For \(\displaystyle -1\le x\le 1\), \[\int\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx = 2\int\tan^{-1}x\,dx = 2\left[x\tan^{-1}x-\frac12\log\left(1+x^{2}\right)\right]+\mathrm{C}\] using Question 13. Final answer: \(\displaystyle 2x\tan^{-1}x-\log\left(1+x^{2}\right)+\mathrm{C}\), for \(\displaystyle -1\le x\le 1\). (Outside that range the identity changes to \(\displaystyle \pi-2\tan^{-1}x\) for \(\displaystyle x>1\) and \(\displaystyle -\pi-2\tan^{-1}x\) for \(\displaystyle x<-1\), giving \(\displaystyle \pi x-2x\tan^{-1}x+\log\left(1+x^{2}\right)+\mathrm{C}\) and \(\displaystyle -\pi x-2x\tan^{-1}x+\log\left(1+x^{2}\right)+\mathrm{C}\) respectively.)

This is a substitution, not integration by parts: the factor \(\displaystyle x^{2}\) is (up to a constant) the derivative of the exponent \(\displaystyle x^{3}\). Put \(\displaystyle t=x^{3}\), so \(\displaystyle dt=3x^{2}dx\), i.e. \(\displaystyle x^{2}dx=\dfrac{dt}{3}\). \[\int x^{2}e^{x^{3}}dx = \frac13\int e^{t}\,dt = \frac13 e^{t}+\mathrm{C} = \frac13 e^{x^{3}}+\mathrm{C}\] Check by differentiating: \(\displaystyle \dfrac{d}{dx}\left(\dfrac13 e^{x^{3}}\right)=\dfrac13 e^{x^{3}}\cdot 3x^{2}=x^{2}e^{x^{3}}\). Final answer: option (A), \(\displaystyle \dfrac13 e^{x^{3}}+\mathrm{C}\).

Expand the bracket to expose the standard form \(\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}\): \[\int e^{x}\sec x(1+\tan x)\,dx = \int e^{x}\left(\sec x+\sec x\tan x\right)dx\] With \(\displaystyle f(x)=\sec x\), \(\displaystyle f'(x)=\sec x\tan x\), so the integrand is exactly \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\). \[= e^{x}\sec x+\mathrm{C}\] Check by differentiating: \(\displaystyle \dfrac{d}{dx}\left(e^{x}\sec x\right)=e^{x}\sec x+e^{x}\sec x\tan x=e^{x}\sec x(1+\tan x)\). Final answer: option (B), \(\displaystyle e^{x}\sec x+\mathrm{C}\).