Question

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

• A. 0
• B. $\pi /2$
• C. 1
• D. $\pi /4$

Answer 1

The correct option is A 0

$I={\int }_{-\pi /2}^{\pi /2}{e}^{{\sin }^{2}x}{\sin }^{(2n+1)}\left(x\right)dx$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-\pi /2}^{\pi /2}{e}^{{\sin }^2(-x)}{\sin }^{(2n+1)}(-x)dx$
$=-{\int }_{-\pi /2}^{\pi /2}{e}^{{\sin }^{2}x}{\sin }^{(2n+1)}xdx=-I\therefore 2I=0\Rightarrow I=0$