Question

The value of the following definite integral is:
${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{sin2x}{1+cosx}dx$

• A. $-2$ In $2$
• B. $2$
• C. $0$
• D. None of these

Answer 1

The correct option is C $0$

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{sin2x}{1+cosx}dx$
$f\left(x\right)=\frac{sin2x}{1+cosx}$
$f(-x)=\frac{sin2(-x)}{1+cos(-x)}=-\frac{sin2x}{1+cosx}$
$=-f\left(x\right)$
$\Rightarrow f\left(x\right)$ is an odd function
$\Rightarrow I=0$