Question

If $f$ is an odd function, then the value of ${\int }_{-a}^a\frac{f\left(\sin x\right)}{f\left(\cos x\right)+f\left({\sin }^{2}x\right)}dx$, is equal to

• A. 0
• B. $f\left(\cos x\right)+f\left(\sin x\right)$
• C. 1
• D. none of these

Answer 1

The correct option is A 0

$I={\int }_{-a}^a\frac{f\left(\sin x\right)}{f\left(\cos x\right)+f\left({\sin }^{2}x\right)}dx$
Let $g\left(x\right)=\frac{f\left(\sin x\right)}{f\left(\cos x\right)+f\left({\sin }^{2}x\right)}$
Then ,$g(-x)=\frac{f\left(\sin \right(-x\left)\right)}{f\left(\cos (-x)\right)+f\left({\sin }^2(-x)\right)}$
$\Rightarrow g(-x)=\frac{f(-sinx)}{f\left(cosx\right)+f\left(si{n}^{2}x\right)}$
$\Rightarrow g(-x)=-\frac{f\left(sinx\right)}{f\left(cosx\right)+f\left(si{n}^{2}x\right)}$
$\Rightarrow g(-x)=-g\left(x\right)$
Hence, g(x) is odd .
So, $I={\int }_{-a}^a\frac{f\left(\sin x\right)}{f\left(\cos x\right)+f\left({\sin }^{2}x\right)}dx=0$