Question

The value of the integral ${\int }_0^{\pi /2}\frac{\sin x+\cos x}{e^{x-\frac{\pi }{4}}+1}dx$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$1$

The given integral

$I={\int }_0^{\pi /2}\frac{\sin x+\cos x}{e^{x-\frac{\pi }{4}}+1}dx$

$={\int }_0^{\pi /2}\frac{\sin (\frac{\pi }{2}-x)+\cos (\frac{\pi }{2}-x)}{e^{\frac{\pi }{2}-x-\frac{\pi }{4}}+1}dx$
$={\int }_0^{\pi /2}\frac{\cos x+\sin x}{e^{-(x-\frac{\pi }{4})}+1}dx$
i.e., $I={\int }_0^{\pi /2}\frac{e^{(x-\frac{\pi }{4})}(\sin x+\cos x)}{1+e^{x-\frac{\pi }{4}}}$
Hence $2I={\int }_0^{\pi /2}\frac{\sin x+\cos x}{e^{x-\frac{\pi }{4}}+1}dx+{\int }_0^{\pi /2}\frac{e^{x-\frac{\pi }{4}}(\sin x+\cos x)}{e^{x-\frac{\pi }{4}}+1}dx$
$={\int }_0^{\pi /2}(\sin x+\cos x)dx$
$=(-\cos x+\sin x{)}_0^{\pi /2}=2$
$\Rightarrow I=1$