JEE Main On 08 April 2018 Question 18
Question: The value of $ \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{{{\sin }^{2}}x}{1+2^{x}}}dx $ is: [JEE Main Online 08-04-2018]
Options:
A) $ 4\pi $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{8} $
D) $ \frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2l=\int _{-\pi |2}^{x/2}{{{\sin }^{2}}xdx} $
As $ {{\sin }^{2}}x $ is even function so,
$ 2l=2\int_0^{\pi /2}{{{\sin }^{2}}xdx} $
$ l=\int_0^{\pi /2}{{{\sin }^{2}}xdx=\frac{\pi }{4}} $
