Definite Integration Question 367
Question: $ \int_0^{\pi }{\frac{dx}{1+\sin x}}= $
[CEE 1993]
Options:
A) 0
B) $ \frac{1}{2} $
C) 2
D) $ \frac{3}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{\pi }{\frac{dx}{1+\sin x}}=\int_0^{\pi }{\frac{1-\sin x}{{{\cos }^{2}}x}dx=\int_0^{\pi }{({{\sec }^{2}}x-\sec x\tan x)dx}} $
$ =[\tan x-\sec x]_0^{\pi }=[\tan \pi -\sec \pi +1]=[0+1+1]=2 $ .
BETA
