Question

The value of $\underset{{}_{-\frac{\pi }{2}}}{\overset{\frac{\pi }{2_{}}}{\int }}$$\frac{cosx}{1+e^x}dx$ is

• A. 2
• B. 0
• C. 1
• D. -2

Answer 1

Answer: (C)

$I=\underset{-{\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos x}{1+e^x}}{\int }dx\dots$(1)
$\Rightarrow I=\underset{-{\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos \left(\frac{\pi }{2}-\frac{\pi }{2}-x\right)}{1+e^{\frac{\pi }{2}-\frac{\pi }{2}-x}}dx}{\int }$
$\Rightarrow I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x}{1+e^{-x}}dx$
$=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{e^x\cos x}{1+e^x}dx\cdots$(2)
From (1) and (2), we have
$2I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{\left(e^x+1\right)\cos x}{e^x+1}dx$
$=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\cos xdx$
$\Rightarrow 2I={\sin x|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$
$\Rightarrow I=1$