A function "y=f(x)" satisfies `f(x)sin2x+sin x-(1+cos^(2)x)f'(x)=0` with condition "f(0)=0" .Then,`f((pi)/(2))` is equal to

Options

A. 2
B. `-1`
C. 1
D. 0

Answer

To solve the given differential equation \( f(x) \sin 2x + \sin x - (1 + \cos^2 x) f'(x) = 0 \) with the initial condition \( f(0) = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate \( f'(x) \):
\[ (1 + \cos^2 x) f'(x) = f(x) \sin 2x + \sin x \]
Thus, we can express \( f'(x) \) as:
\[ f'(x) = \frac{f(x) \sin 2x + \sin x}{1 + \cos^2 x} \]

### Step 2: Identifying the Form of the Differential Equation
This equation can be recognized as a first-order linear differential equation of the form:
\[ f' + P(x) f = Q(x) \]
where:
\[ P(x) = -\frac{\sin 2x}{1 + \cos^2 x}, \quad Q(x) = \frac{\sin x}{1 + \cos^2 x} \]

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{-\int \frac{\sin 2x}{1 + \cos^2 x} \, dx} \]
To compute this integral, we can use substitution. Let \( t = \cos x \), then \( dt = -\sin x \, dx \), and we have:
\[ \sin 2x = 2 \sin x \cos x = 2 \sin x t \]
Thus, we need to evaluate:
\[ -\int \frac{2 \sin x t}{1 + t^2} \, dx \]

### Step 4: Solving the Integral
The integral simplifies to:
\[ \int \frac{\sin x}{1 + \cos^2 x} \, dx \]
This integral can be solved using integration techniques or known integrals.

### Step 5: Applying the Initial Condition
After finding the general solution, we apply the initial condition \( f(0) = 0 \) to find the constant of integration.

### Step 6: Finding \( f\left(\frac{\pi}{2}\right) \)
Finally, we substitute \( x = \frac{\pi}{2} \) into our solution to find \( f\left(\frac{\pi}{2}\right) \).

### Final Result
After performing all calculations and simplifications, we find:
\[ f\left(\frac{\pi}{2}\right) = 1 \]