`int_(0)^(pi) x sin^(4) x dx` is equal to

Options

A. `(3pi)/16`
B. `(3pi^(2))/16`
C. `(16pi)/3`
D. `(16pi^(2))/3`

Explanation

To solve the integral \( I = \int_0^{\pi} x \sin^4 x \, dx \), we can use the property of definite integrals that states:

\[ \int_0^A f(x) \, dx = \int_0^A f(A - x) \, dx \]

### Step 1: Set up the integral
Let \( I = \int_0^{\pi} x \sin^4 x \, dx \).

### Step 2: Use the property of definite integrals
Using the property mentioned, we can express \( I \) as:

\[ I = \int_0^{\pi} (\pi - x) \sin^4(\pi - x) \, dx \]

### Step 3: Simplify the sine function
Using the identity \( \sin(\pi - x) = \sin x \), we have:

\[ I = \int_0^{\pi} (\pi - x) \sin^4 x \, dx \]

### Step 4: Expand the integral
Now we can expand this integral:

\[ I = \int_0^{\pi} \pi \sin^4 x \, dx - \int_0^{\pi} x \sin^4 x \, dx \]

### Step 5: Combine the integrals
Let \( J = \int_0^{\pi} \sin^4 x \, dx \). Then we can express \( I \) as:

\[ I = \pi J - I \]

### Step 6: Solve for \( I \)
Rearranging gives:

\[ 2I = \pi J \implies I = \frac{\pi J}{2} \]

### Step 7: Calculate \( J \)
Now we need to calculate \( J = \int_0^{\pi} \sin^4 x \, dx \). We can use the identity:

\[ \sin^4 x = \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \]

Using \( \cos^2(2x) = \frac{1 + \cos(4x)}{2} \):

\[ \sin^4 x = \frac{1}{4} \left(1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right) = \frac{1}{4} \left(\frac{3}{2} - 2\cos(2x) + \frac{1}{2}\cos(4x)\right) \]

### Step 8: Integrate \( J \)
Now we can integrate:

\[ J = \int_0^{\pi} \sin^4 x \, dx = \frac{1}{4} \left(\frac{3}{2}\int_0^{\pi} 1 \, dx - 2\int_0^{\pi} \cos(2x) \, dx + \frac{1}{2}\int_0^{\pi} \cos(4x) \, dx\right) \]

Calculating each integral:

1. \( \int_0^{\pi} 1 \, dx = \pi \)
2. \( \int_0^{\pi} \cos(2x) \, dx = 0 \) (since it completes a full cycle)
3. \( \int_0^{\pi} \cos(4x) \, dx = 0 \) (same reason)

Thus:

\[ J = \frac{1}{4} \left(\frac{3}{2} \pi - 0 + 0\right) = \frac{3\pi}{8} \]

### Step 9: Substitute back to find \( I \)
Now substituting back into the equation for \( I \):

\[ I = \frac{\pi J}{2} = \frac{\pi \cdot \frac{3\pi}{8}}{2} = \frac{3\pi^2}{16} \]

### Final Answer
Thus, the value of the integral \( \int_0^{\pi} x \sin^4 x \, dx \) is:

\[ \boxed{\frac{3\pi^2}{16}} \]