Question
Evaluate the following integrals:
$\int\frac{\text{x}^2\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}\text{dx}$
$\int\frac{\text{x}^2\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}\text{dx}$
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Answer
Let $\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x dx}}{(1-\text{x}^2)^{\frac{3}{2}}}$
Putting $\text{x}=\sin\theta$
$\Rightarrow\text{dx}=\cos\theta\text{d}\theta$
$\&\ \theta=\sin^{-1}\text{x}$
$\therefore\text{I}=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(1-\sin^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(\cos^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{\cos^3\theta}$
$=\int\tan^2\theta.\theta.\text{d}\theta$
$=\int(\sec^2\theta-1)\theta.\text{d}\theta$.
$=\int\theta.\sec^2\theta\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\int\sec^2\theta\text{d}\theta-\int\Big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\tan\theta-\int1.\tan\theta\text{d}\theta-\frac{\theta^{\ 2}}{2}$
$=\theta.\tan\theta-\ln\big|\sec\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\big|\cos\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\Big|\sqrt{1-\sin^2\theta}\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\theta.\sin\theta}{\sqrt{1-\sin^2\theta}}+\frac{1}2\ln\Big|1-\sin^2\theta\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\ln\big(1-\text{x}^2\big)-\frac{1}{2}\big(\sin^{-1}\text{x}\big)^2+\text{C}$$\Big[\because\theta=\sin^{-1}\text{x}\Big]$
Putting $\text{x}=\sin\theta$
$\Rightarrow\text{dx}=\cos\theta\text{d}\theta$
$\&\ \theta=\sin^{-1}\text{x}$
$\therefore\text{I}=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(1-\sin^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(\cos^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{\cos^3\theta}$
$=\int\tan^2\theta.\theta.\text{d}\theta$
$=\int(\sec^2\theta-1)\theta.\text{d}\theta$.
$=\int\theta.\sec^2\theta\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\int\sec^2\theta\text{d}\theta-\int\Big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\tan\theta-\int1.\tan\theta\text{d}\theta-\frac{\theta^{\ 2}}{2}$
$=\theta.\tan\theta-\ln\big|\sec\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\big|\cos\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\Big|\sqrt{1-\sin^2\theta}\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\theta.\sin\theta}{\sqrt{1-\sin^2\theta}}+\frac{1}2\ln\Big|1-\sin^2\theta\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\ln\big(1-\text{x}^2\big)-\frac{1}{2}\big(\sin^{-1}\text{x}\big)^2+\text{C}$$\Big[\because\theta=\sin^{-1}\text{x}\Big]$
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