Definite Integration Question 7
Question: $ \int _{-\pi /2}^{\pi /2}{\frac{\cos x}{1+e^{x}}dx=} $
[EAMCET 1992]
Options:
A) 1
B) 0
C) $ -1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int _{-\pi /2}^{\pi /2}{\frac{\cos x}{1+e^{x}}dx=\int _{-\pi /2}^{0}{\frac{\cos x}{1+e^{x}}dx+\int_0^{\pi /2}{\frac{\cos x}{1+e^{x}}dx}}} $ …..(i) Putting $ x=-t $ in $ \int _{-\pi /2}^{0}{\frac{\cos x}{1+e^{x}}dx} $ , we get $ I=\int _{-\pi /2}^{0}{\frac{\cos x}{1+e^{x}}dx=\int_0^{\pi /2}{\frac{e^{x}\cos x}{1+e^{x}}dx}} $
$ I=\int_0^{\pi /2}{\frac{e^{x}\cos x}{1+e^{x}}dx+\int_0^{\pi /2}{\frac{\cos x}{1+e^{x}}dx}} $
$ =\int _{0}^{\pi /2}{\frac{(1+e^{x})\cos xdx}{(1+e^{x})}} $
$ =\int_0^{\pi /2}{\cos xdx=[\sin x]_0^{\pi /2}=1} $ .
BETA
