Question

If f is an odd function, then $I={\int }_{-a}^a\frac{f\left(\sin \theta \right)}{f\left(\cos \theta \right)+f\left({\sin }^2\theta \right)}=$

Answer 1

Let $I={\int }_{-a}^a\frac{f\left(\sin \theta \right)}{f\left(\cos \theta \right)+f\left({\sin }^2\theta \right)}d\theta$
And $f\left(\theta \right)=\frac{f\left(\sin \theta \right)}{f\left(\cos \theta \right)+f\left({\sin }^2\theta \right)}$
$f(-\theta )=\frac{f\left(\sin (-\theta )\right)}{f\left(\cos (-\theta )\right)+f\left({\sin }^2(-\theta )\right)}=\frac{f(-\sin \theta )}{f\left(\cos \theta \right)+f\left({\sin }^2\theta \right)}=-f\left(\theta \right)$
As $f$ is odd function
$\therefore I=0$