Definite Integration Question 38
Question: $ \int _{-\pi /2}^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}{e^{-{{\cos }^{2}}x}}dx} $ is equal to
[AMU 1999]
Options:
A) $ 2{e^{-1}} $
B) 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ I=\int _{-\pi /2}^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}{e^{-{{\cos }^{2}}x}}dx} $
$ \because \frac{\sin x}{1+{{\cos }^{2}}x}{e^{-{{\cos }^{2}}x}} $ is an odd function,$ I=0 $ .
BETA
