If f(x) is a continuous function in `[0,pi]` such that f(0)=f(x)=0, then the value of
`int_(0)^(pi//2) {f(2x)-f''(2x)}sin x cos x dx` is equal to

Options

A. `pi`
B. `2pi`
C. `3pi`
D. 0

Explanation

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \left( f(2x) - f''(2x) \right) \sin x \cos x \, dx, \]

given that \( f(0) = f(x) = 0 \) for \( x \in [0, \pi] \), we will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \):

\[ I = \int_{0}^{\frac{\pi}{2}} \left( f(2x) - f''(2x) \right) \sin x \cos x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left( f(2x) - f''(2x) \right) \sin(2x) \, dx. \]

### Step 2: Change of Variables
Let \( t = 2x \). Then, \( dt = 2dx \) or \( dx = \frac{dt}{2} \). The limits change as follows: when \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{2} \), \( t = \pi \). Thus, we have:

\[ I = \frac{1}{2} \int_{0}^{\pi} \left( f(t) - f''(t) \right) \sin(t) \frac{dt}{2} = \frac{1}{4} \int_{0}^{\pi} \left( f(t) - f''(t) \right) \sin(t) \, dt. \]

### Step 3: Split the Integral
Now, we can split the integral into two parts:

\[ I = \frac{1}{4} \left( \int_{0}^{\pi} f(t) \sin(t) \, dt - \int_{0}^{\pi} f''(t) \sin(t) \, dt \right). \]

### Step 4: Integrate by Parts
For the second integral, we will use integration by parts. Let \( u = \sin(t) \) and \( dv = f''(t) dt \). Then, \( du = \cos(t) dt \) and \( v = f'(t) \):

\[ \int f''(t) \sin(t) \, dt = f'(t) \sin(t) \bigg|_{0}^{\pi} - \int f'(t) \cos(t) \, dt. \]

Evaluating the boundary terms, we have:

- At \( t = 0 \): \( f'(0) \sin(0) = 0 \).
- At \( t = \pi \): \( f'(\pi) \sin(\pi) = 0 \).

Thus, the boundary terms contribute \( 0 \), and we have:

\[ \int_{0}^{\pi} f''(t) \sin(t) \, dt = - \int_{0}^{\pi} f'(t) \cos(t) \, dt. \]

### Step 5: Apply Integration by Parts Again
Now we apply integration by parts again to \( \int f'(t) \cos(t) \, dt \):

Let \( u = \cos(t) \) and \( dv = f'(t) dt \). Then, \( du = -\sin(t) dt \) and \( v = f(t) \):

\[ \int f'(t) \cos(t) \, dt = f(t) \cos(t) \bigg|_{0}^{\pi} + \int f(t) \sin(t) \, dt. \]

Again, evaluating the boundary terms:

- At \( t = 0 \): \( f(0) \cos(0) = f(0) = 0 \).
- At \( t = \pi \): \( f(\pi) \cos(\pi) = -f(\pi) \).

Thus, we have:

\[ \int_{0}^{\pi} f'(t) \cos(t) \, dt = -f(\pi) + \int_{0}^{\pi} f(t) \sin(t) \, dt. \]

### Step 6: Combine Everything
Substituting back, we find:

\[ \int_{0}^{\pi} f''(t) \sin(t) \, dt = -\left(-f(\pi) + \int_{0}^{\pi} f(t) \sin(t) \, dt\right) = f(\pi) - \int_{0}^{\pi} f(t) \sin(t) \, dt. \]

Now substituting this back into our expression for \( I \):

\[ I = \frac{1}{4} \left( \int_{0}^{\pi} f(t) \sin(t) \, dt - (f(\pi) - \int_{0}^{\pi} f(t) \sin(t) \, dt) \right). \]

This simplifies to:

\[ I = \frac{1}{4} \left( 2 \int_{0}^{\pi} f(t) \sin(t) \, dt - f(\pi) \right). \]

### Step 7: Evaluate the Integral
Given that \( f(0) = f(\pi) = 0 \), the integral \( \int_{0}^{\pi} f(t) \sin(t) \, dt \) will also yield \( 0 \) if \( f(t) \) is continuous and satisfies the boundary conditions. Therefore, we conclude:

\[ I = \frac{1}{4} \left( 0 - 0 \right) = 0. \]

### Final Answer
Thus, the value of the integral is

\[ \boxed{0}. \]