Question

${\int }_{-\pi /2}^{\pi /2}\frac{{\pi }^{\sin x}}{1+{\pi }^{\sin x}}dx$

• A. $0$
• B. $\pi /4$
• C. $\pi /2$
• D. $\pi$

Answer 1

The correct option is C $\pi /2$

Let $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\pi }^{sinx}}{1+{\pi }^{sinx}}dx$ ...(1)
Using property ${\int }_A^{B}f\left(x\right)dx={\int }_A^{B}f(A+B-x)dx$
$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\pi }^{sin(-x)}}{1+{\pi }^{sin(-x)}}dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{1+{\pi }^{sinx}}dx$ ...(2)
Adding (1) and (2)
$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1dx={\left[x\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=(\frac{\pi }{2}+\frac{\pi }{2})=\pi \Rightarrow I=\frac{\pi }{2}$