Definite Integration Question 87
Question: $ \int_0^{\pi }{\frac{xdx}{1+\sin x}} $ is equal to
[UPSEAT 2004]
Options:
A) $ -\pi $
B) $ \frac{\pi }{2} $
C) $ \pi $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let, $ I=\int_0^{\pi }{\frac{xdx}{1+\sin x}} $ …..(i) $ I=\int_0^{\pi }{\frac{(\pi -x)dx}{1+\sin (\pi -x)}} $
$ I=\int_0^{\pi }{\frac{(\pi -x)dx}{1+\sin x}} $ ….. (ii), $ { \because \int_0^{a}{f(x)dx}=\int_0^{a}{f(a-x)dx} } $
Adding (i) and (ii), we get $ 2I=\int_0^{\pi }{\frac{\pi dx}{1+\sin x}} $
$ 2I=\pi \int_0^{\pi }{\frac{1-\sin x}{(1+\sin x)(1-\sin x)}dx} $
$ 2I=\pi \int_0^{\pi }{\frac{1-\sin x}{{{\cos }^{2}}x}}dx=\pi \int_0^{\pi }{({{\sec }^{2}}x-\sec x\tan x)dx} $
$ 2I=\pi [\tan x-\sec x]_0^{\pi }=\pi [0-(-1)-(0-1)] $ , $ 2I=2\pi $
$ I=\pi $ .B
BETA
