`int_(0)^(pi//2) sin ^(4) x dx`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx \), we can use the power-reduction formula for sine. Here’s a step-by-step solution:

### Step 1: Use the power-reduction formula
The power-reduction formula for \( \sin^2 x \) is:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
To find \( \sin^4 x \), we square \( \sin^2 x \):
\[ \sin^4 x = \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 \]
Expanding this gives:
\[ \sin^4 x = \frac{(1 - \cos(2x))^2}{4} = \frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \]

### Step 2: Substitute \( \cos^2(2x) \) using the power-reduction formula
We can further simplify \( \cos^2(2x) \) using the power-reduction formula:
\[ \cos^2(2x) = \frac{1 + \cos(4x)}{2} \]
Substituting this into our expression for \( \sin^4 x \):
\[ \sin^4 x = \frac{1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}}{4} \]
This simplifies to:
\[ \sin^4 x = \frac{1 - 2\cos(2x) + \frac{1}{2} + \frac{\cos(4x)}{2}}{4} = \frac{3/2 - 2\cos(2x) + \frac{1}{2}\cos(4x)}{4} \]
Thus,
\[ \sin^4 x = \frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x) \]

### Step 3: Set up the integral
Now we can set up the integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx = \int_{0}^{\frac{\pi}{2}} \left(\frac{3}{8} - \frac{1}{2}\cos(2x) + \frac{1}{8}\cos(4x)\right) dx \]

### Step 4: Integrate term by term
Now we integrate each term separately:
1. For \( \frac{3}{8} \):
\[ \int_{0}^{\frac{\pi}{2}} \frac{3}{8} \, dx = \frac{3}{8} \cdot \left[\frac{\pi}{2} - 0\right] = \frac{3\pi}{16} \]

2. For \( -\frac{1}{2}\cos(2x) \):
\[ \int_{0}^{\frac{\pi}{2}} -\frac{1}{2}\cos(2x) \, dx = -\frac{1}{2} \cdot \left[\frac{\sin(2x)}{2}\right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \cdot \left[0 - 0\right] = 0 \]

3. For \( \frac{1}{8}\cos(4x) \):
\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{8}\cos(4x) \, dx = \frac{1}{8} \cdot \left[\frac{\sin(4x)}{4}\right]_{0}^{\frac{\pi}{2}} = \frac{1}{8} \cdot \left[0 - 0\right] = 0 \]

### Step 5: Combine the results
Combining the results from the integrals:
\[ \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx = \frac{3\pi}{16} + 0 + 0 = \frac{3\pi}{16} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx = \frac{3\pi}{16} \]