`int_(0)^(pi//2)sin^(6)x cos^(2)x dx=`

Options

A. `(5)/(256)`
B. `(9pi)/(16)`
C. `(5pi)/(256)`
D. `(pi)/(4)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^6 x \cos^2 x \, dx \), we will use the formula derived from Wallis's theorem for integrals of the form \( \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx \).

### Step-by-Step Solution:

1. **Identify m and n**:
In our case, we have \( m = 6 \) and \( n = 2 \).

2. **Check the conditions for Wallis's theorem**:
According to Wallis's theorem, if \( m \) is even and \( n \) is even, we can use the formula:
\[ \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2} \]
Here, \( !! \) denotes the double factorial.

3. **Calculate the double factorials**:
- For \( m = 6 \):
\[ (m-1)!! = 5!! = 5 \times 3 \times 1 = 15 \]
- For \( n = 2 \):
\[ (n-1)!! = 1!! = 1 \]
- For \( m+n = 8 \):
\[ (m+n)!! = 8!! = 8 \times 6 \times 4 \times 2 = 384 \]

4. **Substitute into the formula**:
Now substituting the values into the formula:
\[ I = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2} = \frac{15 \cdot 1}{384} \cdot \frac{\pi}{2} \]

5. **Simplify the expression**:
\[ I = \frac{15 \pi}{768} \]

6. **Final answer**:
Thus, the value of the integral is:
\[ I = \frac{15 \pi}{768} \]