Evaluate the following integrals: ^ -1 x x^2 dx - Vidyadip

Question

Evaluate the following integrals:$\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$
$=\int\Big(\frac{1}{\text{x}^2}\Big)(\sin^{-1}\text{x})\text{dx}$
$\text{I}=\Big[\sin^{-1}\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big(\frac{1}{\sqrt{1-\text{x}^2}}\int\frac{1}{\text{x}^2}\text{dx}\Big)\text{dx}\Big]$
$=\sin^{-1}\text{x}\big(-\frac{1}{\text{x}}\big)-\int\frac{1}{\sqrt{1-\text{x}^2}}\Big(-\frac{1}{\text{x}}\Big)\text{dx}$
$\text{I} =-\frac{1}{\text{x}}\sin^{-1}\text{x}+\int\frac{1}{\text{x}\sqrt{1-\text{x}^2}}\text{dx}$
$\text{I}=-\frac{1}{\text{x}}\sin^{-1}\text{x}+\text{I}_1 \dots(1)$
Where,
$\text{I}_1=\int\frac{1}{\text{x}\sqrt{1-\text{x}^2}}\text{dx}$
Let $1-\text{x}^2=\text{t}^2$
$-2\text{x dx}=2\text{t dt}$
$\text{I}_1=\int\frac{\text{x}}{\text{x}^2\sqrt{1-\text{x}^2}}\text{dx}$
$=-\int\frac{\text{tdt}}{(1-\text{t}^2)\sqrt{\text{t}}}$
$=-\int\frac{\text{dt}}{(1-\text{t}^2)}$
$=\int\frac{1}{\text{t}^2-1}\text{dt}$
$=\frac{1}{2}\log\Big|\frac{\text{t}-1}{\text{t+1}}\Big|$
$=\frac{1}2{\log}\Big|\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2}+1}\Big|+\text{C}_1$
Now
$\text{I}=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}{2}\log\bigg|\Big(\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2+1}}\Big)\Big(\frac{\sqrt{1-\text{x}^2}-1}{\sqrt{1-\text{x}^2-1}}\Big)\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}2{}\log\bigg|\frac{(\sqrt{1-\text{x}^2}-1)^2}{1-\text{x}^2-1}\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\frac{1}{2}\log\bigg|\frac{(\sqrt{1-\text{x}^2}-1)^2}{-\text{x}^2}\bigg|+\text{C}$
$=-\frac{\sin^{-1}\text{x}}{\text{x}}+\log\bigg|\frac{\sqrt{1-\text{x}^2}-1}{-\text{x}}\bigg|+\text{C}$
$\text{I}=-\frac{\sin^{-1}\text{x}}{\text{x}}+\log\bigg|\frac{1-\sqrt{1-\text{x}}^2}{\text{x}}\bigg|+\text{C}$

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