# If f is an odd function then $$\int_{-a}^{a}\frac{f(sin x)}{f(cos x)+ f(sin^2x)}dx$$

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1. Does not exist
2. π/2
3. π/4
4. 0

Option 4 : 0
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Electric charges and coulomb's law (Basic)
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## Detailed Solution

Formula used:

$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$$

sin (-x) = - sin x

cos (-x) = cos x

Calculation:

Let

$$I = \int_{-a}^{a}\frac{f(sin x)}{f(cos x)+ f(sin^2x)}dx$$      ----(1)

$$I = \int_{-a}^{a}\frac{f[sin (-x)]}{f[cos (-x)]+ [sin^2(-x)]}dx$$

$$I = \int_{-a}^{a}\frac{f(-sin x)}{f(cos x)+ f(sin^2x)}dx$$

Since f is an odd function. Therefore, f(- x) = - f(x)

$$I = -\int_{-a}^{a}\frac{f(sin x)}{f(cos x)+ f(sin^2x)}dx$$      ----(2)

Equation (1) + (2)

$$2I = \int_{-a}^{a}\frac{f(sin x)}{f(cos x)+ f(sin^2x)}dx-\int_{-a}^{a}\frac{f(sin x)}{f(cos x)+ f(sin^2x)}dx$$

2I = 0

I = 0

∴ $$\int_{-a}^{a}\frac{f(sin x)}{f(cos x)+ f(sin^2x)}dx$$ = 0
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