Evaluate :
`int_(0)^(pi//2)sin^(9)x cos^(7)xdx`

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^9 x \cos^7 x \, dx \), we can use the Beta function or the formula for integrals of the form \( \int_{0}^{\frac{\pi}{2}} \sin^{m} x \cos^{n} x \, dx \).

### Step-by-Step Solution:

1. **Identify the parameters**:
Here, we have \( m = 9 \) and \( n = 7 \).

2. **Use the formula**:
The integral can be evaluated using the formula:
\[ \int_{0}^{\frac{\pi}{2}} \sin^{m} x \cos^{n} x \, dx = \frac{1}{2} B\left(\frac{m+1}{2}, \frac{n+1}{2}\right) \]
where \( B(x, y) \) is the Beta function defined as:
\[ B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt \]

3. **Calculate the Beta function**:
For our case:
\[ B\left(\frac{9+1}{2}, \frac{7+1}{2}\right) = B(5, 4) \]

4. **Use the relationship between Beta and Gamma functions**:
The Beta function can also be expressed in terms of Gamma functions:
\[ B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} \]
Therefore:
\[ B(5, 4) = \frac{\Gamma(5) \Gamma(4)}{\Gamma(9)} \]

5. **Calculate the Gamma values**:
We know:
\[ \Gamma(5) = 4! = 24, \quad \Gamma(4) = 3! = 6, \quad \text{and} \quad \Gamma(9) = 8! = 40320 \]

6. **Substitute the values into the Beta function**:
\[ B(5, 4) = \frac{24 \times 6}{40320} = \frac{144}{40320} = \frac{1}{280} \]

7. **Final calculation**:
Now substituting back into the integral:
\[ I = \frac{1}{2} B(5, 4) = \frac{1}{2} \cdot \frac{1}{280} = \frac{1}{560} \]

### Final Answer:
\[ I = \frac{1}{560} \]