NCERT Solutions · Class 12 Mathematics Integrals

Integration by parts: \(\displaystyle \displaystyle\int u\,v\,dx = u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx\). By the ILATE order the algebraic factor \(\displaystyle x\) is the first function and \(\displaystyle \sin x\) the second. \[\int x\sin x\,dx = x(-\cos x)-\int 1\cdot(-\cos x)\,dx = -x\cos x+\int\cos x\,dx\] \[= -x\cos x+\sin x+\mathrm{C}\] Final answer: \(\displaystyle -x\cos x+\sin x+\mathrm{C}\).

Integration by parts with first function \(\displaystyle x\) and second function \(\displaystyle \sin 3x\); note \(\displaystyle \displaystyle\int\sin 3x\,dx=-\frac{\cos 3x}{3}\) (the \(\displaystyle \tfrac13\) from the chain rule is the step most often dropped). \[\int x\sin 3x\,dx = x\left(-\frac{\cos 3x}{3}\right)-\int 1\cdot\left(-\frac{\cos 3x}{3}\right)dx\] \[= -\frac{x\cos 3x}{3}+\frac13\int\cos 3x\,dx = -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}+\mathrm{C}\] Final answer: \(\displaystyle -\dfrac{x\cos 3x}{3}+\dfrac{\sin 3x}{9}+\mathrm{C}\).

Integration by parts, first function \(\displaystyle x^{2}\), second function \(\displaystyle e^{x}\). The parts have to be applied twice, because one application only lowers the power of \(\displaystyle x\) by one. \[\int x^{2}e^{x}\,dx = x^{2}e^{x}-\int 2x\,e^{x}\,dx\] For the new integral, parts again with first function \(\displaystyle x\): \[\int x e^{x}\,dx = x e^{x}-\int e^{x}\,dx = x e^{x}-e^{x}\] Hence \[\int x^{2}e^{x}\,dx = x^{2}e^{x}-2\left(x e^{x}-e^{x}\right)+\mathrm{C} = e^{x}\left(x^{2}-2x+2\right)+\mathrm{C}\] Final answer: \(\displaystyle e^{x}\left(x^{2}-2x+2\right)+\mathrm{C}\).

Here the logarithm has no elementary antiderivative but an easy derivative, so by ILATE \(\displaystyle \log x\) is the FIRST function and \(\displaystyle x\) the second. \[\int x\log x\,dx = \log x\cdot\frac{x^{2}}{2}-\int\frac{1}{x}\cdot\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}\log x-\frac12\int x\,dx\] \[= \frac{x^{2}}{2}\log x-\frac{x^{2}}{4}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{x^{2}}{2}\log x-\dfrac{x^{2}}{4}+\mathrm{C}\), valid for \(\displaystyle x>0\).

Take \(\displaystyle \log 2x\) as the first function and \(\displaystyle x\) as the second. The key point: \(\displaystyle \dfrac{d}{dx}\log 2x=\dfrac{1}{2x}\cdot 2=\dfrac{1}{x}\) — the same derivative as \(\displaystyle \log x\), since \(\displaystyle \log 2x=\log 2+\log x\). \[\int x\log 2x\,dx = \log 2x\cdot\frac{x^{2}}{2}-\int\frac{1}{x}\cdot\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}\log 2x-\frac12\int x\,dx\] \[= \frac{x^{2}}{2}\log 2x-\frac{x^{2}}{4}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{x^{2}}{2}\log 2x-\dfrac{x^{2}}{4}+\mathrm{C}\), valid for \(\displaystyle x>0\).

By ILATE, \(\displaystyle \log x\) is the first function and \(\displaystyle x^{2}\) the second. \[\int x^{2}\log x\,dx = \log x\cdot\frac{x^{3}}{3}-\int\frac{1}{x}\cdot\frac{x^{3}}{3}\,dx = \frac{x^{3}}{3}\log x-\frac13\int x^{2}\,dx\] \[= \frac{x^{3}}{3}\log x-\frac{x^{3}}{9}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{x^{3}}{3}\log x-\dfrac{x^{3}}{9}+\mathrm{C}\), valid for \(\displaystyle x>0\).

By ILATE the inverse-trigonometric factor \(\displaystyle \sin^{-1}x\) is the first function, \(\displaystyle x\) the second, and \(\displaystyle \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}}\). \[\int x\sin^{-1}x\,dx = \frac{x^{2}}{2}\sin^{-1}x-\frac12\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx\] For the remaining integral write \(\displaystyle x^{2}=-\left(1-x^{2}\right)+1\), which splits it into two standard forms: \[\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx = \int\frac{dx}{\sqrt{1-x^{2}}}-\int\sqrt{1-x^{2}}\,dx = \sin^{-1}x-\left[\frac{x}{2}\sqrt{1-x^{2}}+\frac12\sin^{-1}x\right]\] \[= \frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}\] Substituting back, \[\int x\sin^{-1}x\,dx = \frac{x^{2}}{2}\sin^{-1}x-\frac14\sin^{-1}x+\frac{x}{4}\sqrt{1-x^{2}}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{2x^{2}-1}{4}\sin^{-1}x+\dfrac{x\sqrt{1-x^{2}}}{4}+\mathrm{C}\), for \(\displaystyle -1\le x\le 1\).

By ILATE, \(\displaystyle \tan^{-1}x\) is the first function and \(\displaystyle x\) the second; \(\displaystyle \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}}\). \[\int x\tan^{-1}x\,dx = \frac{x^{2}}{2}\tan^{-1}x-\frac12\int\frac{x^{2}}{1+x^{2}}\,dx\] The fraction is improper, so divide before integrating: \(\displaystyle \dfrac{x^{2}}{1+x^{2}}=1-\dfrac{1}{1+x^{2}}\). \[\int\frac{x^{2}}{1+x^{2}}\,dx = \int\left(1-\frac{1}{1+x^{2}}\right)dx = x-\tan^{-1}x\] Hence \[\int x\tan^{-1}x\,dx = \frac{x^{2}}{2}\tan^{-1}x-\frac{x}{2}+\frac12\tan^{-1}x+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{x^{2}+1}{2}\tan^{-1}x-\dfrac{x}{2}+\mathrm{C}\).

By ILATE, \(\displaystyle \cos^{-1}x\) is the first function and \(\displaystyle x\) the second. The sign is the trap here: \(\displaystyle \dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}\). \[\int x\cos^{-1}x\,dx = \frac{x^{2}}{2}\cos^{-1}x-\int\left(-\frac{1}{\sqrt{1-x^{2}}}\right)\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}\cos^{-1}x+\frac12\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx\] Writing \(\displaystyle x^{2}=1-\left(1-x^{2}\right)\), \[\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx = \sin^{-1}x-\left[\frac{x}{2}\sqrt{1-x^{2}}+\frac12\sin^{-1}x\right] = \frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}\] So \[\int x\cos^{-1}x\,dx = \frac{x^{2}}{2}\cos^{-1}x+\frac14\sin^{-1}x-\frac{x}{4}\sqrt{1-x^{2}}+\mathrm{C}\] Using \(\displaystyle \sin^{-1}x=\dfrac{\pi}{2}-\cos^{-1}x\) and absorbing the constant \(\displaystyle \dfrac{\pi}{8}\) into \(\displaystyle \mathrm{C}\): \[= \frac{2x^{2}-1}{4}\cos^{-1}x-\frac{x}{4}\sqrt{1-x^{2}}+\mathrm{C}\] Final answer: \(\displaystyle \dfrac{2x^{2}-1}{4}\cos^{-1}x-\dfrac{x\sqrt{1-x^{2}}}{4}+\mathrm{C}\), for \(\displaystyle -1\le x\le 1\).

Substitute \(\displaystyle x=\sin\theta\), so \(\displaystyle \theta=\sin^{-1}x\) with \(\displaystyle \theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) and \(\displaystyle dx=\cos\theta\,d\theta\). \[I=\int\left(\sin^{-1}x\right)^{2}dx = \int\theta^{2}\cos\theta\,d\theta\] Parts with first function \(\displaystyle \theta^{2}\): \[\int\theta^{2}\cos\theta\,d\theta = \theta^{2}\sin\theta-\int 2\theta\sin\theta\,d\theta\] Parts again on \(\displaystyle \displaystyle\int\theta\sin\theta\,d\theta=-\theta\cos\theta+\sin\theta\), so \[I = \theta^{2}\sin\theta-2\left(-\theta\cos\theta+\sin\theta\right)+\mathrm{C} = \theta^{2}\sin\theta+2\theta\cos\theta-2\sin\theta+\mathrm{C}\] On \(\displaystyle \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\) the cosine is non-negative, so \(\displaystyle \cos\theta=+\sqrt{1-x^{2}}\) (this positive root is the step to be careful about). Putting \(\displaystyle \theta=\sin^{-1}x\), \(\displaystyle \sin\theta=x\): \[I = x\left(\sin^{-1}x\right)^{2}+2\sqrt{1-x^{2}}\,\sin^{-1}x-2x+\mathrm{C}\] Final answer: \(\displaystyle x\left(\sin^{-1}x\right)^{2}+2\sqrt{1-x^{2}}\,\sin^{-1}x-2x+\mathrm{C}\), for \(\displaystyle -1\le x\le 1\).