Given `f(x) = sin^(3) x` and p(x) is a quadratic polynomial with leading coefficient unity. Then the value of `int_(0)^(2pi)p(x)f''(x)dx` is _____

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_{0}^{2\pi} p(x) f''(x) \, dx \]

where \( f(x) = \sin^3 x \) and \( p(x) \) is a quadratic polynomial with leading coefficient 1.

### Step 1: Determine \( f''(x) \)

First, we need to find the second derivative \( f''(x) \) of \( f(x) = \sin^3 x \).

Using the chain rule and product rule, we first find \( f'(x) \):

\[ f'(x) = 3 \sin^2 x \cos x \]

Now, we differentiate \( f'(x) \) to find \( f''(x) \):

\[ f''(x) = 3 \left( 2 \sin x \cos x \cos x + \sin^2 x (-\sin x) \right) \]
\[ = 3 \left( 2 \sin x \cos^2 x - \sin^3 x \right) \]
\[ = 3 \sin x (2 \cos^2 x - \sin^2 x) \]

### Step 2: Set up the integral

Now, we substitute \( f''(x) \) into the integral:

\[ I = \int_{0}^{2\pi} p(x) f''(x) \, dx \]

Assuming \( p(x) = x^2 + bx + c \) (since it is a quadratic polynomial with leading coefficient 1), we have:

\[ I = \int_{0}^{2\pi} (x^2 + bx + c) \cdot 3 \sin x (2 \cos^2 x - \sin^2 x) \, dx \]

### Step 3: Analyze the integral

Notice that \( \sin x \) and \( \cos^2 x \) are periodic functions with a period of \( 2\pi \). The integral of any odd function over a complete period (from \( 0 \) to \( 2\pi \)) is zero.

1. The term \( \sin x \) is an odd function.
2. The term \( x^2 \) is even, \( bx \) is odd, and \( c \) is even.

Thus, when we multiply \( p(x) \) (which contains both even and odd terms) by \( \sin x \) (which is odd), the integral of the odd terms will cancel out over the interval \( [0, 2\pi] \).

### Step 4: Evaluate the integral

Since the integral of an odd function over a symmetric interval around zero is zero, we conclude:

\[ I = 0 \]

### Final Answer

The value of the integral

\[ \int_{0}^{2\pi} p(x) f''(x) \, dx = 0 \]