Question

Integrate ${\int }_{-\pi /2}^{\pi /2}{e}^{{\sin }^{-2}x}.{\sin }^{2n+1}xdx=$

Answer 1

Let $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{e}^{{\left({\sin }^{-1}x\right)}^2}{\sin }^{2n+1}xdx$

since $f\left(x\right)=e^{{\left({\sin }^{-1}x\right)}^2}{\sin }^{2n+1}xdx$ is an odd function

So, $I=0$