Question

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

• A. 2
• B. 0
• C. 1
• D. -2

Answer 1

Answer: (C)

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