Definite Integration Question 230
Question: $ \int _{0}^{1}{\sin ( 2{{\tan }^{-1}}\sqrt{\frac{1+x}{1-x}} )dx=} $
[EAMCET 2003]
Options:
A) $ \pi /6 $
B) $ \pi /4 $
C) $ \pi /2 $
D) $ \pi $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_0^{1}{\sin ( 2{{\tan }^{-1}}\sqrt{\frac{1+x}{1-x}} )dx} $
Put $ x=\cos \theta , $ then $ \sin [ 2{{\tan }^{-1}}\sqrt{\frac{1+\cos \theta }{1-\cos \theta }} ] $
$ =\sin [ 2{{\tan }^{-1}}( \cot \frac{\theta }{2} ) ] $
$ =\sin [ 2{{\tan }^{-1}}[ \tan ( \frac{\pi }{2}-\frac{\theta }{2} ) ] ]=\sin [ 2( \frac{\pi }{2}-\frac{\theta }{2} ) ] $
$ =\sin (\pi -\theta )=\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-x^{2}} $
Now, $ \int_0^{1}{\sin ( 2{{\tan }^{-1}}\sqrt{\frac{1+x}{1-x}} )}dx=\int_0^{1}{\sqrt{1-x^{2}}dx} $
$ =[ \frac{1}{2}x\sqrt{1-x^{2}} ]_0^{1}+\frac{1}{2}[{{\sin }^{-1}}x]_0^{1}=\frac{\pi }{4}. $
BETA
