Evaluate the following: ^ -1 x (1-x^2)^ 3 4 dx - Vidyadip

Question

Evaluate the following:
$\int\frac{\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{4}}}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{4}}}\text{dx}$ $=\int\frac{\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}\text{dx}$
Put $\sin^{-1}\text{x}=\text{t}\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
And $\text{x}-\sin\text{t}\Rightarrow1-\text{x}^2=\cos^2\text{t}$
$\cos\text{t}=\sqrt{1-\text{x}^2}$
$\text{I}=\int\frac{\text{t}}{\cos^2\text{t}}\text{dt}=\int\text{t}\cdot\sec^2\text{tdt}$
$=\text{t}\cdot\int\sec^2\text{tdt}-\int\Big(\frac{\text{d}}{\text{dt}}\text{t}\cdot\int\sec^2\text{tdt}\Big)\text{dt}$
$=\text{t}\cdot\tan\text{t}-\int1\cdot\tan\text{tdt}$
$=\text{t}\tan\text{t}+\log|\cos\text{t}|+\text{C}$ $\Big[\because\int\tan\text{xdx}=-\log|\cos\text{x}|+\text{C}\Big]$
$=\sin^{-1}\text{x}\cdot\frac{\text{x}}{\sqrt{1-\text{x}^2}}+\log\Big|\sqrt{1-\text{x}^2}\Big|+\text{C}$

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