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

Question

Evaluate the following integrals:
$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$

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Answer

$\int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}$
$=\int\sin^{-1}(\sin2\text{x} )\text{dx}$ $\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big]$
$=\int2\text{ x dx}$
$=2\int\text{x dx}$
$=\frac{2\text{x}^2}{2}+\text{C}$
$=\text{x}^2+\text{C}$
$\therefore\ \int\sin^{-1}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)\text{dx}=\text{x}^2+\text{C}$

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