NCERT Solutions · Class 12 Mathematics Integrals

The factor \(\displaystyle (x+1)\) is repeated, so it needs two partial-fraction terms: \[\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2}, \] so \(\displaystyle x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}\). Putting \(\displaystyle x=-1\): \(\displaystyle 1=B\). Putting \(\displaystyle x=-2\): \(\displaystyle 3=C\). Comparing coefficients of \(\displaystyle x^{2}\): \(\displaystyle 1=A+C\), so \(\displaystyle A=-2\). \[\int\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}dx=-2\int\frac{dx}{x+1}+\int\frac{dx}{(x+1)^{2}}+3\int\frac{dx}{x+2} \] \[=-2\log|x+1|-\frac{1}{x+1}+3\log|x+2|+\mathrm{C}=\log\left|\frac{(x+2)^{3}}{(x+1)^{2}}\right|-\frac{1}{x+1}+\mathrm{C}. \]

Simplify the inverse function first. Put \(\displaystyle x=\cos 2\theta\), so \(\displaystyle \theta=\frac12\cos^{-1}x\); then by \(\displaystyle 1-\cos 2\theta=2\sin^{2}\theta\) and \(\displaystyle 1+\cos 2\theta=2\cos^{2}\theta\), \[\tan^{-1}\sqrt{\frac{1-x}{1+x}}=\tan^{-1}\sqrt{\tan^{2}\theta}=\theta=\frac12\cos^{-1}x,\qquad -1<x\le 1, \] since \(\displaystyle \theta\in\left[0,\frac{\pi}{2}\right)\) there, where \(\displaystyle \tan\theta\ge 0\). So the integral is \(\displaystyle \frac12\displaystyle\int\cos^{-1}x\,dx\). Integrating by parts with \(\displaystyle \cos^{-1}x\) as the first function and \(\displaystyle 1\) as the second, \[\int\cos^{-1}x\,dx=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^{2}}}dx=x\cos^{-1}x-\sqrt{1-x^{2}}. \] Hence \[\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx=\frac12\left[x\cos^{-1}x-\sqrt{1-x^{2}}\right]+\mathrm{C}. \]

Combine the logarithms: \(\displaystyle \log\left(x^{2}+1\right)-2\log x=\log\dfrac{x^{2}+1}{x^{2}}=\log\left(1+\dfrac{1}{x^{2}}\right)\). Also, for \(\displaystyle x>0\), \(\displaystyle \dfrac{\sqrt{x^{2}+1}}{x^{4}}=\dfrac{x\sqrt{1+1/x^{2}}}{x^{4}}=\dfrac{1}{x^{3}}\sqrt{1+\dfrac{1}{x^{2}}}\). So the integrand is \[\frac{1}{x^{3}}\sqrt{1+\frac{1}{x^{2}}}\;\log\left(1+\frac{1}{x^{2}}\right). \] Put \(\displaystyle t=1+\dfrac{1}{x^{2}}\), so \(\displaystyle dt=-\dfrac{2}{x^{3}}dx\), i.e. \(\displaystyle \dfrac{dx}{x^{3}}=-\dfrac{dt}{2}\): \[I=-\frac12\int t^{1/2}\log t\,dt. \] Integrate by parts with \(\displaystyle \log t\) as the first function: \[\int t^{1/2}\log t\,dt=\frac23 t^{3/2}\log t-\frac23\int t^{1/2}dt=\frac23 t^{3/2}\log t-\frac49 t^{3/2}=\frac23 t^{3/2}\left(\log t-\frac23\right). \] Therefore \[I=-\frac13 t^{3/2}\left(\log t-\frac23\right)+\mathrm{C}=-\frac13\left(1+\frac{1}{x^{2}}\right)^{3/2}\left[\log\left(1+\frac{1}{x^{2}}\right)-\frac23\right]+\mathrm{C}. \]

Use the half-angle forms \(\displaystyle 1-\cos x=2\sin^{2}\frac{x}{2}\) and \(\displaystyle \sin x=2\sin\frac{x}{2}\cos\frac{x}{2}\): \[\frac{1-\sin x}{1-\cos x}=\frac{1-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\sin^{2}\dfrac{x}{2}}=\frac12\,\mathrm{cosec}^{2}\frac{x}{2}-\cot\frac{x}{2}. \] Take \(\displaystyle f(x)=-\cot\frac{x}{2}\); then \(\displaystyle f'(x)=\frac12\mathrm{cosec}^{2}\frac{x}{2}\), so the integrand is \(\displaystyle e^{x}\left[f(x)+f'(x)\right]\) and \[\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)dx=e^{x}f(x)+\mathrm{C}=-e^{x}\cot\frac{x}{2}+\mathrm{C}. \] Evaluating between the limits (note \(\displaystyle \cot\frac{\pi}{2}=0\) and \(\displaystyle \cot\frac{\pi}{4}=1\)), \[\int_{\pi/2}^{\pi}e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)dx=\left[-e^{x}\cot\frac{x}{2}\right]_{\pi/2}^{\pi}=-e^{\pi}\cdot 0+e^{\pi/2}\cdot 1=e^{\pi/2}. \]

Divide numerator and denominator by \(\displaystyle \cos^{4}x\); this is legitimate on \(\displaystyle \left[0,\frac{\pi}{4}\right]\), where \(\displaystyle \cos x\neq 0\): \[\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}=\frac{\tan x\sec^{2}x}{1+\tan^{4}x}. \] Put \(\displaystyle t=\tan^{2}x\), so \(\displaystyle dt=2\tan x\sec^{2}x\,dx\). The limits change: \(\displaystyle x=0\Rightarrow t=0\) and \(\displaystyle x=\frac{\pi}{4}\Rightarrow t=1\). Hence \[\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}dx=\frac12\int_{0}^{1}\frac{dt}{1+t^{2}}=\frac12\left[\tan^{-1}t\right]_{0}^{1}=\frac12\cdot\frac{\pi}{4}=\frac{\pi}{8}. \]

Write the numerator in terms of the denominator. Using \(\displaystyle 1=\cos^{2}x+\sin^{2}x\), look for constants \(\displaystyle A,B\) with \[\cos^{2}x=A\left(\cos^{2}x+4\sin^{2}x\right)+B\left(\cos^{2}x+\sin^{2}x\right). \] Comparing coefficients of \(\displaystyle \cos^{2}x\): \(\displaystyle 1=A+B\); of \(\displaystyle \sin^{2}x\): \(\displaystyle 0=4A+B\). So \(\displaystyle A=-\frac13\), \(\displaystyle B=\frac43\), and \[\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}=-\frac13+\frac43\cdot\frac{1}{\cos^{2}x+4\sin^{2}x}=-\frac13+\frac43\cdot\frac{\sec^{2}x}{1+4\tan^{2}x}. \] For the second piece put \(\displaystyle t=\tan x\), \(\displaystyle dt=\sec^{2}x\,dx\); as \(\displaystyle x\to\frac{\pi}{2}^{-}\) we have \(\displaystyle t\to\infty\), so \[\frac43\int_{0}^{\infty}\frac{dt}{1+4t^{2}}=\frac43\cdot\frac12\left[\tan^{-1}2t\right]_{0}^{\infty}=\frac23\cdot\frac{\pi}{2}=\frac{\pi}{3}. \] Therefore \[\int_{0}^{\pi/2}\frac{\cos^{2}x\,dx}{\cos^{2}x+4\sin^{2}x}=-\frac13\cdot\frac{\pi}{2}+\frac{\pi}{3}=-\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6}. \]

The numerator \(\displaystyle \sin x+\cos x\) is the derivative of \(\displaystyle \sin x-\cos x\), and \(\displaystyle \sin 2x\) can be expressed through that same quantity: \[(\sin x-\cos x)^{2}=1-2\sin x\cos x=1-\sin 2x\quad\Longrightarrow\quad \sin 2x=1-(\sin x-\cos x)^{2}. \] Put \(\displaystyle t=\sin x-\cos x\), so \(\displaystyle dt=(\cos x+\sin x)\,dx\). The limits become \[x=\frac{\pi}{6}\Rightarrow t=\frac12-\frac{\sqrt3}{2}=-\frac{\sqrt3-1}{2},\qquad x=\frac{\pi}{3}\Rightarrow t=\frac{\sqrt3}{2}-\frac12=\frac{\sqrt3-1}{2}. \] Hence \[\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx=\int_{-\frac{\sqrt3-1}{2}}^{\frac{\sqrt3-1}{2}}\frac{dt}{\sqrt{1-t^{2}}}=\left[\sin^{-1}t\right]_{-\frac{\sqrt3-1}{2}}^{\frac{\sqrt3-1}{2}}. \] Since \(\displaystyle \sin^{-1}\) is an odd function, \[=2\sin^{-1}\left(\frac{\sqrt3-1}{2}\right). \]

Rationalise the denominator; its conjugate reduces it to \(\displaystyle (1+x)-x=1\): \[\frac{1}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x}=\sqrt{1+x}+\sqrt{x}. \] Hence \[\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}=\int_{0}^{1}\left(\sqrt{1+x}+\sqrt{x}\right)dx=\left[\frac23 (1+x)^{3/2}+\frac23 x^{3/2}\right]_{0}^{1} \] \[=\left(\frac23\cdot 2\sqrt2+\frac23\right)-\left(\frac23+0\right)=\frac{4\sqrt2}{3}. \]

Again \(\displaystyle \sin x+\cos x\) is the derivative of \(\displaystyle \sin x-\cos x\). Put \(\displaystyle t=\sin x-\cos x\), so \(\displaystyle dt=(\sin x+\cos x)\,dx\) and \(\displaystyle \sin 2x=1-t^{2}\). The limits become \(\displaystyle x=0\Rightarrow t=-1\) and \(\displaystyle x=\frac{\pi}{4}\Rightarrow t=0\), and the denominator becomes \(\displaystyle 9+16\left(1-t^{2}\right)=25-16t^{2}\): \[I=\int_{-1}^{0}\frac{dt}{25-16t^{2}}=\frac{1}{16}\int_{-1}^{0}\frac{dt}{\left(\frac54\right)^{2}-t^{2}}. \] Using \(\displaystyle \int\frac{dt}{a^{2}-t^{2}}=\frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|\) with \(\displaystyle a=\frac54\), \[I=\frac{1}{16}\cdot\frac{2}{5}\left[\log\left|\frac{5+4t}{5-4t}\right|\right]_{-1}^{0}=\frac{1}{40}\left(\log 1-\log\frac19\right)=\frac{1}{40}\log 9=\frac{1}{20}\log 3. \]

Put \(\displaystyle t=\sin x\), so \(\displaystyle dt=\cos x\,dx\) and \(\displaystyle \sin 2x\,dx=2\sin x\cos x\,dx=2t\,dt\); the limits become \(\displaystyle 0\) and \(\displaystyle 1\): \[I=\int_{0}^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx=\int_{0}^{1}2t\tan^{-1}t\,dt. \] Integrate by parts with \(\displaystyle \tan^{-1}t\) as the first function and \(\displaystyle 2t\) as the second: \[I=\left[t^{2}\tan^{-1}t\right]_{0}^{1}-\int_{0}^{1}\frac{t^{2}}{1+t^{2}}dt=\frac{\pi}{4}-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right)dt, \] where \(\displaystyle \dfrac{t^{2}}{1+t^{2}}\) was made proper before integrating. Hence \[I=\frac{\pi}{4}-\left[t-\tan^{-1}t\right]_{0}^{1}=\frac{\pi}{4}-\left(1-\frac{\pi}{4}\right)=\frac{\pi}{2}-1. \]