`int_(0)^(2pi) sin^(5) (x/4)` dx is equal to

Options

A. `8/15`
B. `16/15`
C. `32/15`
D. `0`

Explanation

To solve the integral \( \int_{0}^{2\pi} \sin^5\left(\frac{x}{4}\right) \, dx \), we can follow these steps:

### Step 1: Use the identity for sine squared
We can express \( \sin^5\left(\frac{x}{4}\right) \) in terms of \( \sin^2\left(\frac{x}{4}\right) \):
\[ \sin^5\left(\frac{x}{4}\right) = \sin^2\left(\frac{x}{4}\right) \cdot \sin^3\left(\frac{x}{4}\right) \]
Using the identity \( \sin^2\theta = 1 - \cos^2\theta \), we can write:
\[ \sin^2\left(\frac{x}{4}\right) = 1 - \cos^2\left(\frac{x}{4}\right) \]

### Step 2: Rewrite the integral
Now, we can rewrite the integral:
\[ \int_{0}^{2\pi} \sin^5\left(\frac{x}{4}\right) \, dx = \int_{0}^{2\pi} \left(1 - \cos^2\left(\frac{x}{4}\right)\right) \sin^3\left(\frac{x}{4}\right) \, dx \]

### Step 3: Use substitution
Let \( t = \cos\left(\frac{x}{4}\right) \). Then, we find \( dt = -\frac{1}{4} \sin\left(\frac{x}{4}\right) \, dx \) or \( dx = -4 \frac{dt}{\sin\left(\frac{x}{4}\right)} \).

### Step 4: Change the limits of integration
When \( x = 0 \), \( t = \cos(0) = 1 \) and when \( x = 2\pi \), \( t = \cos\left(\frac{2\pi}{4}\right) = \cos(\frac{\pi}{2}) = 0 \). Thus, the limits change from \( 0 \) to \( 2\pi \) to \( 1 \) to \( 0 \).

### Step 5: Substitute and integrate
Substituting into the integral:
\[ \int_{1}^{0} \left(1 - t^2\right) \left(-4\right) dt = 4 \int_{0}^{1} \left(1 - t^2\right) dt \]
Now, calculate the integral:
\[ 4 \int_{0}^{1} (1 - t^2) dt = 4 \left[ t - \frac{t^3}{3} \right]_{0}^{1} = 4 \left( 1 - \frac{1}{3} \right) = 4 \cdot \frac{2}{3} = \frac{8}{3} \]

### Step 6: Final result
Thus, the value of the integral is:
\[ \int_{0}^{2\pi} \sin^5\left(\frac{x}{4}\right) \, dx = \frac{8}{3} \]