`int_(0)^(pi) sin^(5) ((x)/(2))dx` equals

Options

A. `(16)/(15`
B. `(32)/(15)`
C. `(8)/(15)`
D. `(5)/(6)`

Answer

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

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int_{0}^{\pi} \sin^5\left(\frac{x}{2}\right)dx = \int_{0}^{\pi} \sin^4\left(\frac{x}{2}\right) \sin\left(\frac{x}{2}\right)dx \]

### Step 2: Use the Identity for Sine Squared
Next, we use the identity \( \sin^2\theta = 1 - \cos^2\theta \):
\[ \sin^4\left(\frac{x}{2}\right) = \left(\sin^2\left(\frac{x}{2}\right)\right)^2 = \left(1 - \cos^2\left(\frac{x}{2}\right)\right)^2 \]
This gives us:
\[ I = \int_{0}^{\pi} \left(1 - \cos^2\left(\frac{x}{2}\right)\right)^2 \sin\left(\frac{x}{2}\right)dx \]

### Step 3: Substitute for Cosine
Let \( t = \cos\left(\frac{x}{2}\right) \). Then, we differentiate:
\[ dt = -\frac{1}{2}\sin\left(\frac{x}{2}\right)dx \implies dx = -2\frac{dt}{\sin\left(\frac{x}{2}\right)} \]
The limits change as follows:
- When \( x = 0 \), \( t = \cos(0) = 1 \)
- When \( x = \pi \), \( t = \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, we can rewrite the integral:
\[ I = \int_{1}^{0} \left(1 - t^2\right)^2 (-2)dt = 2 \int_{0}^{1} \left(1 - t^2\right)^2 dt \]

### Step 4: Expand the Integrand
Now we expand \( (1 - t^2)^2 \):
\[ (1 - t^2)^2 = 1 - 2t^2 + t^4 \]
So, the integral becomes:
\[ I = 2 \int_{0}^{1} \left(1 - 2t^2 + t^4\right) dt \]

### Step 5: Integrate Term by Term
Now we integrate term by term:
\[ I = 2 \left( \int_{0}^{1} 1 dt - 2 \int_{0}^{1} t^2 dt + \int_{0}^{1} t^4 dt \right) \]
Calculating each integral:
- \( \int_{0}^{1} 1 dt = 1 \)
- \( \int_{0}^{1} t^2 dt = \frac{t^3}{3} \Big|_{0}^{1} = \frac{1}{3} \)
- \( \int_{0}^{1} t^4 dt = \frac{t^5}{5} \Big|_{0}^{1} = \frac{1}{5} \)

Putting it all together:
\[ I = 2 \left( 1 - 2 \cdot \frac{1}{3} + \frac{1}{5} \right) \]

### Step 6: Simplify the Expression
Now simplify the expression inside the parentheses:
\[ 1 - \frac{2}{3} + \frac{1}{5} = \frac{15}{15} - \frac{10}{15} + \frac{3}{15} = \frac{8}{15} \]
Thus,
\[ I = 2 \cdot \frac{8}{15} = \frac{16}{15} \]

### Final Answer
The value of the integral is:
\[ \boxed{\frac{16}{15}} \]