Question

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

• A. does not exist
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $0$

Answer 1

Answer: (D)

$0$

Let $I={\int }_{-a}^a\frac{f\left(\sin x\right)}{f\left(\cos x\right)+f\left(si{n}^{2}x\right)}dx$ ...(1)
Using ${\int }_a^{b}f\left(x\right)={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-a}^a\frac{f\left(\sin (-x)\right)}{f\left(\cos (-x)\right)+f\left(si{n}^2(-x)\right)}dx$
$={\int }_{-a}^a\frac{f(-\sin x)}{f\left(\cos (-x)\right)+f\left(si{n}^{2}x\right)}dx$
$=-{\int }_{-a}^a\frac{f\left(\sin x\right)}{f\left(\cos x\right)+f\left(si{n}^{2}x\right)}dx$ ...(2)
From (1) and (2), we get
$2I=0\Rightarrow I=0$