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

Question

Evaluate the following integrals:

$\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$

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Answer

Let $\text{I}=\int\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)\text{dx}$
Let $\text{x}=\tan\text{t}$
$\text{dx}=\sec^2\text{t dt}$
$\text{I}=\int\cos^{-1}\Big(\frac{1-\tan^2\text{t}}{1+\tan^2\text{t}}\Big)\sec^2\text{t dt}$
$=\int\cos^{-1}(\cos2\text{t})\sec^2\text{t dt}$
$=\int2\text{t}\sec^2\text{x dx}$
$=2\Big[\text{t}\int\sec^2\text{t dt}-\int(1\int\sec^2\text{t dt})\text{dt}\Big]$
$=2[\text{t}\tan^2\text{t}-\int\tan\text{t dt}]$
$=2[\text{t}\tan^2\text{t}-\log\sec\text{t}]+\text{C}$
$=2\Big[\text{x}\tan^{-1}\text{x}-\log\sqrt{1+\text{x}^2}\Big]+\text{C}$
$\text{I}=2\text{x}\tan^{-1}\text{x}-\log|1+\text{x}^2|+\text{C}$

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