If f is an odd function, then `I= int_(-a)^(a) (f (sin theta))/(f (cos theta) +f(sin^(2) theta))=`

Options

A. 0
B. `pi//2`
C. 2
D. None of these

Answer

To solve the integral \( I = \int_{-a}^{a} \frac{f(\sin \theta)}{f(\cos \theta) + f(\sin^2 \theta)} \, d\theta \) given that \( f \) is an odd function, we can follow these steps:

### Step 1: Define the property of odd functions
An odd function \( f(x) \) satisfies the property:
\[ f(-x) = -f(x) \]

### Step 2: Substitute \( \theta \) with \( -\theta \)
To analyze the integral, we can substitute \( \theta \) with \( -\theta \):
\[ I = \int_{-a}^{a} \frac{f(\sin(-\theta))}{f(\cos(-\theta)) + f(\sin^2(-\theta))} \, d\theta \]

### Step 3: Simplify using the properties of sine and cosine
Using the properties of sine and cosine:
- \( \sin(-\theta) = -\sin(\theta) \)
- \( \cos(-\theta) = \cos(\theta) \)

Thus, we can rewrite the integral:
\[ I = \int_{-a}^{a} \frac{f(-\sin \theta)}{f(\cos \theta) + f(\sin^2 \theta)} \, d\theta \]

### Step 4: Apply the property of odd function \( f \)
Since \( f \) is an odd function:
\[ f(-\sin \theta) = -f(\sin \theta) \]
This gives us:
\[ I = \int_{-a}^{a} \frac{-f(\sin \theta)}{f(\cos \theta) + f(\sin^2 \theta)} \, d\theta \]

### Step 5: Factor out the negative sign
We can factor out the negative sign from the integral:
\[ I = -\int_{-a}^{a} \frac{f(\sin \theta)}{f(\cos \theta) + f(\sin^2 \theta)} \, d\theta \]
This implies:
\[ I = -I \]

### Step 6: Solve for \( I \)
Adding \( I \) to both sides gives:
\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \boxed{0} \]