Question

${\int }_0^{\frac{\pi }{2}}\frac{e^{\sin x}}{e^{\sin x}+e^{\cos x}}$ dx $=$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{8}$

Answer 1

The correct option is A $\frac{\pi }{4}$

$I={\int }_0^{\frac{\pi }{2}}\frac{e^{\sin x}}{e^{\sin x}+e^{\cos x}}dx$ ..... (1)

We know that,

${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{e^{\sin (\frac{\pi }{2}-x)}}{e^{\sin (\frac{\pi }{2}-x)}+e^{\cos (\frac{\pi }{2}-x)}}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{e^{\cos x}}{e^{\sin x}+e^{\cos x}}dx$ ........ (2)

$2I={\int }_0^{\frac{\pi }{2}}1\cdot dx$

$I=\frac{1}{2}[x{]}_0^{\frac{\pi }{2}}$

$I=\frac{1}{2}(\frac{\pi }{2}-0)$

$=\frac{\pi }{4}$