A) 0
B) \[\frac{1}{2}\]
C) 2
D) \[\frac{3}{2}\]
Correct Answer: C
Solution :
\[\int_{0}^{\pi }{\frac{dx}{1+\sin x}}=\int_{0}^{\pi }{\frac{1-\sin x}{{{\cos }^{2}}x}dx=\int_{0}^{\pi }{({{\sec }^{2}}x-\sec x\tan x)dx}}\] \[=[\tan x-\sec x]_{0}^{\pi }=[\tan \pi -\sec \pi +1]=[0+1+1]=2\].
You need to login to perform this action.
You will be redirected in
3 sec
