Mathematics

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


ANSWER

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


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
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