NCERT Solutions · Class 12 Mathematics Integrals

Substitution in a definite integral. Put \(\displaystyle t=x^{2}+1\), so that \(\displaystyle dt=2x\,dx\), i.e. \(\displaystyle x\,dx=\tfrac{1}{2}\,dt\). Change the limits along with the variable — \(\displaystyle x=0\Rightarrow t=1\) and \(\displaystyle x=1\Rightarrow t=2\) — so that no back-substitution is needed. \[\int_{0}^{1}\frac{x}{x^{2}+1}\,dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}=\frac{1}{2}\Big[\log|t|\Big]_{1}^{2}=\frac{1}{2}\left(\log 2-\log 1\right)\] Final answer: \(\displaystyle \dfrac{1}{2}\log 2\).

The odd power of \(\displaystyle \cos\phi\) is the cue: keep one factor \(\displaystyle \cos\phi\,d\phi\) for the substitution and convert the rest using \(\displaystyle \cos^{2}\phi=1-\sin^{2}\phi\). Put \(\displaystyle \sin\phi=t\), so \(\displaystyle \cos\phi\,d\phi=dt\). Limits: \(\displaystyle \phi=0\Rightarrow t=0\) and \(\displaystyle \phi=\frac{\pi}{2}\Rightarrow t=1\). \[\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\phi}\,\cos^{5}\phi\,d\phi=\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\phi}\,\left(1-\sin^{2}\phi\right)^{2}\cos\phi\,d\phi=\int_{0}^{1}t^{\frac{1}{2}}\left(1-t^{2}\right)^{2}dt\] Expand before integrating: \[\int_{0}^{1}\left(t^{\frac{1}{2}}-2t^{\frac{5}{2}}+t^{\frac{9}{2}}\right)dt=\left[\frac{2}{3}t^{\frac{3}{2}}-\frac{4}{7}t^{\frac{7}{2}}+\frac{2}{11}t^{\frac{11}{2}}\right]_{0}^{1}=\frac{2}{3}-\frac{4}{7}+\frac{2}{11}\] \[=\frac{154-132+42}{231}=\frac{64}{231}\] Final answer: \(\displaystyle \dfrac{64}{231}\).

Simplify the inverse-trigonometric function first. Put \(\displaystyle x=\tan\theta\), i.e. \(\displaystyle \theta=\tan^{-1}x\); then \[\frac{2x}{1+x^{2}}=\frac{2\tan\theta}{1+\tan^{2}\theta}=\sin 2\theta\] The domain check is the step that decides the answer. As \(\displaystyle x\) runs over \(\displaystyle [0,1]\), \(\displaystyle \theta=\tan^{-1}x\) runs over \(\displaystyle \left[0,\frac{\pi}{4}\right]\), so \(\displaystyle 2\theta\in\left[0,\frac{\pi}{2}\right]\), which lies inside the principal range \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) of \(\displaystyle \sin^{-1}\). Only for that reason may we write \(\displaystyle \sin^{-1}(\sin 2\theta)=2\theta\). Hence on \(\displaystyle [0,1]\) \[\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)=2\tan^{-1}x\] So \(\displaystyle \mathrm{I}=2\displaystyle\int_{0}^{1}\tan^{-1}x\,dx\). Integrate by parts taking \(\displaystyle \tan^{-1}x\) as the first function and \(\displaystyle 1\) as the second: \[\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\frac{x}{1+x^{2}}dx=x\tan^{-1}x-\frac{1}{2}\log\left(1+x^{2}\right)\] Therefore \[\mathrm{I}=2\left[x\tan^{-1}x-\frac{1}{2}\log\left(1+x^{2}\right)\right]_{0}^{1}=2\left[\left(\frac{\pi}{4}-\frac{1}{2}\log 2\right)-\left(0-0\right)\right]\] Final answer: \(\displaystyle \dfrac{\pi}{2}-\log 2\).

Follow the hint. Put \(\displaystyle x+2=t^{2}\), so \(\displaystyle x=t^{2}-2\), \(\displaystyle dx=2t\,dt\), and \(\displaystyle \sqrt{x+2}=t\) (the positive root, since \(\displaystyle t>0\) over the range). The limits change too: \(\displaystyle x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}\), and \(\displaystyle x=2\Rightarrow t^{2}=4\Rightarrow t=2\). \[\int_{0}^{2}x\sqrt{x+2}\,dx=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right)\cdot t\cdot 2t\,dt=2\int_{\sqrt{2}}^{2}\left(t^{4}-2t^{2}\right)dt\] \[=2\left[\frac{t^{5}}{5}-\frac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}\] Use \(\displaystyle \left(\sqrt{2}\right)^{5}=4\sqrt{2}\) and \(\displaystyle \left(\sqrt{2}\right)^{3}=2\sqrt{2}\): \[=2\left[\left(\frac{32}{5}-\frac{16}{3}\right)-\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\right]=2\left[\frac{96-80}{15}-\left(-\frac{8\sqrt{2}}{15}\right)\right]\] \[=2\left[\frac{16}{15}+\frac{8\sqrt{2}}{15}\right]=\frac{32+16\sqrt{2}}{15}\] Final answer: \(\displaystyle \dfrac{16\left(2+\sqrt{2}\right)}{15}\).

The numerator is, up to sign, the derivative of \(\displaystyle \cos x\), so substitute \(\displaystyle \cos x=t\), giving \(\displaystyle -\sin x\,dx=dt\). Limits: \(\displaystyle x=0\Rightarrow t=1\) and \(\displaystyle x=\frac{\pi}{2}\Rightarrow t=0\). \[\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\cos^{2}x}\,dx=-\int_{1}^{0}\frac{dt}{1+t^{2}}=\int_{0}^{1}\frac{dt}{1+t^{2}}\] (the swap of limits absorbs the minus sign). \[=\Big[\tan^{-1}t\Big]_{0}^{1}=\tan^{-1}1-\tan^{-1}0=\frac{\pi}{4}-0\] Final answer: \(\displaystyle \dfrac{\pi}{4}\).

Complete the square in the denominator so the integral takes a standard form. \[x+4-x^{2}=-\left(x^{2}-x-4\right)=-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-4\right]=\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\] This is \(\displaystyle a^{2}-u^{2}\) with \(\displaystyle a=\frac{\sqrt{17}}{2}\) and \(\displaystyle u=x-\frac{1}{2}\), so use \[\int\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\log\left|\frac{a+u}{a-u}\right|+\mathrm{C},\qquad \frac{1}{2a}=\frac{1}{\sqrt{17}}\] Multiplying numerator and denominator inside the logarithm by \(\displaystyle 2\), \[\int_{0}^{2}\frac{dx}{x+4-x^{2}}=\frac{1}{\sqrt{17}}\left[\log\left|\frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}\right|\right]_{0}^{2}\] \[=\frac{1}{\sqrt{17}}\left[\log\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]=\frac{1}{\sqrt{17}}\log\frac{\left(\sqrt{17}+3\right)\left(\sqrt{17}+1\right)}{\left(\sqrt{17}-3\right)\left(\sqrt{17}-1\right)}\] \[=\frac{1}{\sqrt{17}}\log\frac{20+4\sqrt{17}}{20-4\sqrt{17}}=\frac{1}{\sqrt{17}}\log\frac{5+\sqrt{17}}{5-\sqrt{17}}\] The quadratic \(\displaystyle x+4-x^{2}\) is positive throughout \(\displaystyle [0,2]\) (its least value on the interval is \(\displaystyle 2\), at \(\displaystyle x=2\)), so there is no singularity inside the range and the modulus signs may be dropped. Final answer: \(\displaystyle \dfrac{1}{\sqrt{17}}\log\dfrac{5+\sqrt{17}}{5-\sqrt{17}}\), which on rationalising equals \(\displaystyle \dfrac{1}{\sqrt{17}}\log\dfrac{21+5\sqrt{17}}{4}\approx 0.568\).

Complete the square: \(\displaystyle x^{2}+2x+5=(x+1)^{2}+2^{2}\). Put \(\displaystyle u=x+1\), so \(\displaystyle du=dx\); the limits shift to \(\displaystyle x=-1\Rightarrow u=0\) and \(\displaystyle x=1\Rightarrow u=2\). \[\int_{-1}^{1}\frac{dx}{x^{2}+2x+5}=\int_{0}^{2}\frac{du}{u^{2}+2^{2}}\] By the standard result \(\displaystyle \int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+\mathrm{C}\) with \(\displaystyle a=2\), \[=\frac{1}{2}\left[\tan^{-1}\frac{u}{2}\right]_{0}^{2}=\frac{1}{2}\left(\tan^{-1}1-\tan^{-1}0\right)=\frac{1}{2}\cdot\frac{\pi}{4}\] Final answer: \(\displaystyle \dfrac{\pi}{8}\).

Recognise the standard form \(\displaystyle \displaystyle\int e^{ax}\left[a\,f(x)+f'(x)\right]dx=e^{ax}f(x)+\mathrm{C}\). Here \(\displaystyle a=2\). Try \(\displaystyle f(x)=\dfrac{1}{2x}\), so that \(\displaystyle f'(x)=-\dfrac{1}{2x^{2}}\), and check: \[a\,f(x)+f'(x)=2\cdot\frac{1}{2x}-\frac{1}{2x^{2}}=\frac{1}{x}-\frac{1}{2x^{2}}\] which is exactly the bracket in the question. Hence an antiderivative is \(\displaystyle \dfrac{e^{2x}}{2x}\), and by the second fundamental theorem of calculus (the integrand is continuous on \(\displaystyle [1,2]\), since \(\displaystyle x\neq 0\) there), \[\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2x^{2}}\right)e^{2x}dx=\left[\frac{e^{2x}}{2x}\right]_{1}^{2}=\frac{e^{4}}{4}-\frac{e^{2}}{2}\] Final answer: \(\displaystyle \dfrac{e^{4}}{4}-\dfrac{e^{2}}{2}=\dfrac{e^{2}}{4}\left(e^{2}-2\right)\).

On \(\displaystyle \left[\frac{1}{3},1\right]\) we have \(\displaystyle x>0\), so take \(\displaystyle x^{3}\) — not \(\displaystyle x\) — out of the cube root; that is the step that makes the substitution work. \[\left(x-x^{3}\right)^{\frac{1}{3}}=\left[x^{3}\left(\frac{1}{x^{2}}-1\right)\right]^{\frac{1}{3}}=x\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}\] Therefore \[\mathrm{I}=\int_{\frac{1}{3}}^{1}\frac{x\left(\dfrac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{4}}dx=\int_{\frac{1}{3}}^{1}\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}\frac{dx}{x^{3}}\] Put \(\displaystyle t=\dfrac{1}{x^{2}}-1\); then \(\displaystyle dt=-\dfrac{2}{x^{3}}dx\), i.e. \(\displaystyle \dfrac{dx}{x^{3}}=-\dfrac{dt}{2}\). Limits: \(\displaystyle x=\frac{1}{3}\Rightarrow t=9-1=8\), and \(\displaystyle x=1\Rightarrow t=0\). \[\mathrm{I}=-\frac{1}{2}\int_{8}^{0}t^{\frac{1}{3}}dt=\frac{1}{2}\int_{0}^{8}t^{\frac{1}{3}}dt=\frac{1}{2}\cdot\frac{3}{4}\left[t^{\frac{4}{3}}\right]_{0}^{8}=\frac{3}{8}\left(8^{\frac{4}{3}}\right)=\frac{3}{8}\times 16=6\] Final answer: option (A) 6.

Apply the first fundamental theorem of calculus: if \(\displaystyle f(x)=\displaystyle\int_{a}^{x}g(t)\,dt\) with \(\displaystyle g\) continuous, then \(\displaystyle f\) is differentiable and \(\displaystyle f'(x)=g(x)\) — differentiating with respect to the upper limit simply replaces \(\displaystyle t\) by \(\displaystyle x\). Here \(\displaystyle g(t)=t\sin t\), continuous for all \(\displaystyle t\), so \[f'(x)=x\sin x\] The integral need not be evaluated at all; but as a check, integrating by parts gives \(\displaystyle f(x)=\left[\sin t-t\cos t\right]_{0}^{x}=\sin x-x\cos x\), and then \[f'(x)=\cos x-\left(\cos x-x\sin x\right)=x\sin x\] the same result. Final answer: option (B) \(\displaystyle x\sin x\).