Question

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

• A. $0$
• B. $1$
• C. $\frac{p}{4}$
• D. $\frac{4}{e^{\pi }}$

Answer 1

Answer: (A)

$0$

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

Applying the law-

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

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

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

Hence, we get $I=-I$

$⟹2I=0$

Hence, $I=0$