A) \[\pi /6\]
B) \[\pi /4\]
C) \[\pi /2\]
D) \[\pi \]
Correct Answer: B
Solution :
\[\int_{0}^{1}{\sin \left( 2{{\tan }^{-1}}\sqrt{\frac{1+x}{1-x}} \right)\,dx}\] Put \[x=\cos \theta ,\]then \[\sin \left[ 2{{\tan }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }} \right]\] \[=\sin \,\,\left[ 2{{\tan }^{-1}}\left( \cot \frac{\theta }{2} \right) \right]\] \[=\sin \left[ 2{{\tan }^{-1}}\left[ \tan \left( \frac{\pi }{2}-\frac{\theta }{2} \right) \right] \right]=\sin \left[ 2\,\left( \frac{\pi }{2}-\frac{\theta }{2} \right) \right]\] \[=\sin (\pi -\theta )=\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-{{x}^{2}}}\] Now, \[\int_{0}^{1}{\sin \left( 2{{\tan }^{-1}}\sqrt{\frac{1+x}{1-x}} \right)}\,dx=\int_{0}^{1}{\sqrt{1-{{x}^{2}}}\,dx}\] \[=\left[ \frac{1}{2}x\sqrt{1-{{x}^{2}}} \right]_{0}^{1}+\frac{1}{2}[{{\sin }^{-1}}x]_{0}^{1}=\frac{\pi }{4}.\]
You need to login to perform this action.
You will be redirected in
3 sec
