Mathematics

# $\displaystyle \int \cos \left \{ 2\tan ^{-1}\sqrt{\frac{1-x}{1+x}} \right \}dx$ is equal to

$\displaystyle \frac{1}{2}x^{2}+k$

##### SOLUTION
$\displaystyle \int \cos \left \{ 2\tan ^{-1}\sqrt{\frac{1-x}{1+x}} \right \}dx$

Put $x=\cos { \theta }$
$\Rightarrow dx=-\sin { \theta } d\theta$

$=-\int \cos \left\{ 2\tan ^{ -1 } \sqrt { \tan ^{ 2 }{ (\dfrac { \theta }{ 2 } ) } } \right\} \sin { \theta } d\theta$
$=-\int \cos \theta \sin { \theta } d\theta$
$=\int { xdx }$
$\displaystyle=\frac { { x }^{ 2 } }{ 2 } +k$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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