A function `f(x)` satisfies `f(x)=sinx+int_0^xf^(prime)(t)(2sint-sin^2t)dt` is

Options

A. `f((pi)/(6))=1`
B. `g(x)=int_(0)^(x)f(t)` dt is increasing on `(0,pi)`
C. `f(0)=0`
D. f(x) is increasing on `(0,pi)`

Explanation

To solve the problem, we start with the given equation for the function \( f(x) \):

\[ f(x) = \sin x + \int_0^x f'(t) (2 \sin t - \sin^2 t) dt \]

### Step 1: Differentiate both sides with respect to \( x \)

Using the Fundamental Theorem of Calculus and the chain rule, we differentiate the right-hand side:

\[ f'(x) = \cos x + f'(x)(2 \sin x - \sin^2 x) \]

### Step 2: Rearranging the equation

We can rearrange the equation to isolate \( f'(x) \):

\[ f'(x) - f'(x)(2 \sin x - \sin^2 x) = \cos x \]

Factoring out \( f'(x) \):

\[ f'(x) \left(1 - (2 \sin x - \sin^2 x)\right) = \cos x \]

### Step 3: Simplifying the expression

We can simplify the expression inside the parentheses:

\[ 1 - (2 \sin x - \sin^2 x) = 1 - 2 \sin x + \sin^2 x \]

This can be rewritten as:

\[ (1 - \sin x)^2 \]

Thus, we have:

\[ f'(x) (1 - \sin x)^2 = \cos x \]

### Step 4: Solving for \( f'(x) \)

Now, we can solve for \( f'(x) \):

\[ f'(x) = \frac{\cos x}{(1 - \sin x)^2} \]

### Step 5: Integrating to find \( f(x) \)

Next, we integrate \( f'(x) \) to find \( f(x) \):

\[ f(x) = \int \frac{\cos x}{(1 - \sin x)^2} dx \]

Let \( u = 1 - \sin x \), then \( du = -\cos x \, dx \). Therefore, \( dx = -\frac{du}{\cos x} \). The integral becomes:

\[ f(x) = -\int \frac{1}{u^2} du = \frac{1}{u} + C = \frac{1}{1 - \sin x} + C \]

### Step 6: Finding the constant \( C \)

To find \( C \), we use the condition \( f(0) = 0 \):

\[ f(0) = \frac{1}{1 - \sin(0)} + C = \frac{1}{1} + C = 1 + C \]

Setting this equal to 0 gives:

\[ 1 + C = 0 \implies C = -1 \]

### Final Expression for \( f(x) \)

Thus, we have:

\[ f(x) = \frac{1}{1 - \sin x} - 1 = \frac{1 - (1 - \sin x)}{1 - \sin x} = \frac{\sin x}{1 - \sin x} \]

### Conclusion

The function \( f(x) \) is:

\[ f(x) = \frac{\sin x}{1 - \sin x} \]