Question

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

• A. $ \frac{1}{8}({{x}^{2}}-1)+c $
• B. $ \frac{{{x}^{4}}}{4}+c $
• C. $ \frac{x}{2}+c $
• D. $ \frac{x}{4}+c $
• E. $ \frac{{{x}^{2}}}{2}+c $

Answer 1

Answer: (E)

Let $ I=\int{\cos \left\{ 2{{\tan }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}dx $ Put $ x=cos\theta $
$ \therefore $ $ I=\int{\cos \left\{ 2{{\tan }^{-1}}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }} \right\}}dx $
$=\int{\cos \left\{ 2{{\tan }^{-1}}\left( \tan \frac{\theta }{2} \right) \right\}}dx $
$=\int{\cos \theta dx=\int{x}\,}dx=\frac{{{x}^{2}}}{2}+c $