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

Answer

To evaluate the integral \( \int_{0}^{\frac{\pi}{2}} \sin^8 x \, dx \), we will use Wallis's formula, which is particularly useful for integrals of the form \( \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx \) where \( n \) is an even integer.

### Step-by-Step Solution:

1. **Identify the value of \( n \)**:
Here, we have \( n = 8 \).

2. **Apply Wallis's formula**:
Wallis's formula states that:
\[ \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx \]
For even \( n \), this can be simplified to:
\[ \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx = \frac{(n-1)(n-3)(n-5)\cdots(3)(1)}{n(n-2)(n-4)\cdots(4)(2)} \cdot \frac{\pi}{2} \]

3. **Substitute \( n = 8 \) into the formula**:
\[ \int_{0}^{\frac{\pi}{2}} \sin^8 x \, dx = \frac{(8-1)(8-3)(8-5)(8-7)}{8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} \]

4. **Calculate the numerator**:
\[ (8-1) = 7, \quad (8-3) = 5, \quad (8-5) = 3, \quad (8-7) = 1 \]
So, the numerator becomes:
\[ 7 \cdot 5 \cdot 3 \cdot 1 = 105 \]

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

6. **Combine the results**:
Now we substitute back into the formula:
\[ \int_{0}^{\frac{\pi}{2}} \sin^8 x \, dx = \frac{105}{384} \cdot \frac{\pi}{2} \]

7. **Simplify**:
\[ = \frac{105 \pi}{768} \]

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