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) \[\frac{{{x}^{2}}}{2}+c\]
Correct Answer: E
Solution :
Let \[I=\int{\cos \left\{ 2{{\tan }^{-1}}\sqrt{\frac{1-x}{1+x}} \right\}}dx\] Put \[x=cos\theta \] \[\therefore \] \[I=\int{\cos \left\{ 2{{\tan }^{-1}}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }} \right\}}dx\] \[=\int{\cos \left\{ 2{{\tan }^{-1}}\left( \tan \frac{\theta }{2} \right) \right\}}dx\] \[=\int{\cos \theta dx=\int{x}\,}dx=\frac{{{x}^{2}}}{2}+c\]
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