If f(x) is a function satisfying `f((1)/(x))+x^(2)f(x)=0` for all non zero x then, ` int_(sin theta)^("cosec" theta) f(x)dx` equals

Options

A. `sin theta+"cosec" theta`
B. `sin^(2)theta`
C. `"cosec" ^(2)theta`
D. none of these

Explanation

To solve the problem, we need to find the value of the integral

\[ I = \int_{\sin \theta}^{\csc \theta} f(x) \, dx \]

given that \( f\left(\frac{1}{x}\right) + x^2 f(x) = 0 \) for all non-zero \( x \).

### Step-by-step Solution:

1. **Express \( f(x) \) in terms of \( f\left(\frac{1}{x}\right) \)**:
From the equation \( f\left(\frac{1}{x}\right) + x^2 f(x) = 0 \), we can rearrange it to find:
\[ f\left(\frac{1}{x}\right) = -x^2 f(x) \]

2. **Substitute \( f(x) \) into the integral**:
We can substitute the expression for \( f\left(\frac{1}{x}\right) \) into the integral:
\[ I = \int_{\sin \theta}^{\csc \theta} f(x) \, dx = \int_{\sin \theta}^{\csc \theta} -\frac{1}{x^2} f\left(\frac{1}{x}\right) \, dx \]

3. **Change of variable**:
Let \( t = \frac{1}{x} \). Then, \( dx = -\frac{1}{t^2} dt \). When \( x = \sin \theta \), \( t = \csc \theta \), and when \( x = \csc \theta \), \( t = \sin \theta \). Thus, the limits of integration change:
\[ I = \int_{\csc \theta}^{\sin \theta} -\frac{1}{\left(\frac{1}{t}\right)^2} f(t) \left(-\frac{1}{t^2}\right) dt \]
Simplifying gives:
\[ I = \int_{\csc \theta}^{\sin \theta} f(t) \, dt \]

4. **Rearranging the integral**:
Now we can write:
\[ I = \int_{\sin \theta}^{\csc \theta} f(x) \, dx = \int_{\csc \theta}^{\sin \theta} f(t) \, dt \]

5. **Combining the integrals**:
Notice that:
\[ I = -I \]
This implies:
\[ 2I = 0 \implies I = 0 \]

### Final Result:
Thus, the value of the integral is:
\[ \int_{\sin \theta}^{\csc \theta} f(x) \, dx = 0 \]