Definite Integration Question 465
Question: $ \int_0^{\pi }{\frac{x\tan x}{\sec x+\cos x}}dx= $
[MNR 1985; BIT Ranchi 1986; UPSEAT 2002]
Options:
A) $ \frac{{{\pi }^{2}}}{4} $
B) $ \frac{{{\pi }^{2}}}{2} $
C) $ \frac{3{{\pi }^{2}}}{2} $
D) $ \frac{{{\pi }^{2}}}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let I = $ \int_0^{\pi }{\frac{x\tan x}{\sec x+\cos x}dx}=\int_0^{\pi }{\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\cos (\pi -x)}dx} $
It gives $ I=\frac{\pi }{2}\int_0^{\pi }{\frac{\sin x}{1+{{\cos }^{2}}x}}dx $
Now put $ \cos x=t $ and solve, we get $ I=\frac{\pi }{2}\times \frac{\pi }{2}=\frac{{{\pi }^{2}}}{4} $ .
BETA
