Definite Integration Question 464
Question: $ \int_0^{\pi }{\frac{x\tan x}{\sec x+\tan x}}dx= $
[MNR 1984]
Options:
A) $ \frac{\pi }{2}-1 $
B) $ \pi ( \frac{\pi }{2}+1 ) $
C) $ \frac{\pi }{2}+1 $
D) $ \pi ( \frac{\pi }{2}-1 ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_0^{\pi }{\frac{x\tan x}{\sec x+\tan x}dx=\int_0^{\pi }{\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}}dx} $
therefore $ 2I=\frac{\pi }{2}\int_0^{\pi }{\frac{\tan x}{\sec x+\tan x}dx=\frac{\pi }{2}\int_0^{\pi }{\frac{\sin x}{1+\sin x}dx}} $
= $ \frac{\pi }{2}[ \int_0^{\pi }{1dx-\int_0^{\pi }{\frac{dx}{1+\sin x}}} ] $
On solving, we get $ I=\frac{{{\pi }^{2}}}{2}-\pi =\pi ( \frac{\pi }{2}-1 ) $ .
BETA
