If `m gt 0, n gt 0`, the definite integral `l=int_(0)^(1)x^(m-1)(1-x)^(n-1)dx` depends upon the vlaues of m and n and is denoted by `beta(m,n)`, called the beta function. E.g.
`int_(0)^(1)x^(4)(1-x)^(5)dx=int_(0)^(1)x^(5-1)(1-x)^(6-1)dx=beta(5, 6) and int_(0)^(1)x^(5//2)(1-x)^(-1//2)dx=int_(0)^(1)x^(7//2-1)(1-x)^(1//2-1)dx=beta((7)/(2),(1)/(2))`. Obviously, `beta(n, m)=beta(m, n)`.
The integral `int_(0)^(pi//2)cos^(2m)theta sin^(2n) theta d theta` is equal to

Options

A. `(1)/(2) beta (m+(1)/(2), n+(1)/(2))`
B. `2beta(2m, 2n)`
C. `beta(2m+1, 2n+1)`
D. none of these

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \cos^{2m} \theta \sin^{2n} \theta \, d\theta \), we can use the properties of the Beta function and some substitutions.

### Step-by-Step Solution:

1. **Recognize the Integral Form**:
The integral can be expressed in terms of the Beta function:
\[ I = \int_{0}^{\frac{\pi}{2}} \cos^{2m} \theta \sin^{2n} \theta \, d\theta \]

2. **Substitution**:
We can use the substitution \( x = \sin^2 \theta \). Then, \( dx = 2 \sin \theta \cos \theta \, d\theta \) or \( d\theta = \frac{dx}{2 \sqrt{x(1-x)}} \). The limits change from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) which corresponds to \( x = 0 \) to \( x = 1 \).

3. **Change the Integral**:
The integral becomes:
\[ I = \int_{0}^{1} (1-x)^{m-1} x^{n-1} \frac{dx}{2 \sqrt{x(1-x)}} \]
Simplifying, we get:
\[ I = \frac{1}{2} \int_{0}^{1} x^{n-1} (1-x)^{m-1} \, dx \]

4. **Identify the Beta Function**:
The integral \( \int_{0}^{1} x^{n-1} (1-x)^{m-1} \, dx \) is the Beta function \( B(n, m) \):
\[ I = \frac{1}{2} B(n, m) \]

5. **Use the Property of Beta Function**:
The Beta function can also be expressed in terms of Gamma functions:
\[ B(n, m) = \frac{\Gamma(n) \Gamma(m)}{\Gamma(n+m)} \]
Therefore, we have:
\[ I = \frac{1}{2} \cdot \frac{\Gamma(n) \Gamma(m)}{\Gamma(n+m)} \]

6. **Final Result**:
Thus, the value of the integral is:
\[ I = \frac{1}{2} B(n, m) = \frac{1}{2} \cdot \frac{\Gamma(n) \Gamma(m)}{\Gamma(n+m)} \]