Definite Integration Question 347
Question: The value of $ \int _{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+a^{x}}dx,a>0,} $ is
[IIT Screening 2001; AIEEE 2005]
Options:
A) $ \pi $
B) $ a\pi $
C) $ \frac{\pi }{2} $
D) $ 2\pi $
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int _{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+a^{x}}dx}=\int _{\pi }^{-\pi }{\frac{{{\cos }^{2}}x}{1+{a^{-x}}}(-dx)} $ (Putting- x for x)
$ =\int _{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{a^{-x}}}}dx $
$ \Rightarrow I+I=\int _{-\pi }^{\pi }{{{\cos }^{2}}x( \frac{1}{1+a^{x}}+\frac{1}{1+{a^{-x}}} )dx} $
$ =\int _{-\pi }^{\pi }{{{\cos }^{2}}xdx} $
therefore 2I $ =2\int_0^{\pi }{{{\cos }^{2}}x.dx=}\int_0^{\pi }{(1+\cos 2x)dx} $
therefore 2I $ =[x]_0^{\pi }+[ \frac{\sin 2x}{2} ]_0^{\pi } $
$ \Rightarrow 2I=\pi \Rightarrow I=\frac{\pi }{2} $ .
BETA
