Show the t \[ \int_{0}^{\frac{\pi}{2}} \sin ^{2 n-1} \theta \cos ^{2 n-1} \theta d \theta=\frac{\Gamma(m) r(n)}{2 \Gamma(m+n)} \]

To solve the integral \[ I = \int_{0}^{\frac{\pi}{2}} \sin^{2n-1} \theta \cos^{2n-1} \theta \, d\theta, \] we can use the substitution \( u = \sin^2 \theta \). This gives us \( du = 2 \sin \theta \cos \theta \, d\theta \), or \( d\theta = \frac{du}{2\sqrt{u(1-u)}} \). The limits of integration change as follows: - When \( \theta = 0 \), \( u = 0 \). - When \( \theta = \frac{\pi}{2} \), \( u = 1 \). Thus, we can rewrite the integral as: \[ I = \int_{0}^{1} u^{n-1} (1-u)^{n-1} \frac{du}{2\sqrt{u(1-u)}}. \] This integral can be simplified further. The integral can be expressed in terms of the Beta function: \[ I = \frac{1}{2} \int_{0}^{1} u^{n-1} (1-u)^{n-1} \, du = \frac{1}{2} B(n, n), \] where \( B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} \). Thus, we have: \[ I = \frac{1}{2} \cdot \frac{\Gamma(n) \Gamma(n)}{\Gamma(2n)}. \] Now, we can express this in terms of \( \Gamma(m) \) and \( r(n) \). We can set \( m = n \) and \( r(n) = \frac{\Gamma(n)}{2} \). Therefore, we can write: \[ I = \frac{\Gamma(m) r(n)}{2 \Gamma(m+n)}. \] This gives us the desired result: \[ \int_{0}^{\frac{\pi}{2}} \sin^{2n-1} \theta \cos^{2n-1} \theta \, d\theta = \frac{\Gamma(n) \cdot \frac{\Gamma(n)}{2}}{2 \Gamma(2n)}. \] Thus, the final expression is: \[ \int_{0}^{\frac{\pi}{2}} \sin^{2n-1} \theta \cos^{2n-1} \theta \, d\theta = \frac{\Gamma(n) \cdot \frac{\Gamma(n)}{2}}{2 \Gamma(2n)}. \]