Evaluate :
`int_(0)^(pi//2)sin^(8)xdx`

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^8 x \, dx \), we can use the Wallis formula for integrals of sine functions.

### Step-by-Step Solution:

1. **Identify the Integral**:
We need to evaluate:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^8 x \, dx \]

2. **Apply Wallis' Formula**:
Wallis' formula states that for even \( n \):
\[ \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \frac{(n-1)(n-3)(n-5)\ldots(3)(1)}{n(n-2)(n-4)\ldots(4)(2)} \cdot \frac{\pi}{2} \]
Here, \( n = 8 \) (which is even).

3. **Calculate the Numerator**:
The numerator is:
\[ (8-1)(8-3)(8-5)(8-7) = 7 \cdot 5 \cdot 3 \cdot 1 = 105 \]

4. **Calculate the Denominator**:
The denominator is:
\[ 8 \cdot 6 \cdot 4 \cdot 2 = 384 \]

5. **Combine the Results**:
Now substitute these values into Wallis' formula:
\[ I = \frac{105}{384} \cdot \frac{\pi}{2} \]

6. **Final Calculation**:
\[ I = \frac{105 \pi}{768} \]

Thus, the value of the integral \( \int_{0}^{\frac{\pi}{2}} \sin^8 x \, dx \) is:
\[ \boxed{\frac{105 \pi}{768}} \]