2.2 Use the reduction formula in a) to show that: \[ \int_{0}^{\frac{\pi}{2}} \sin ^{n} x d x=\frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin ^{n-2} x d x \]

To show the reduction formula \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx, \] we can start by using integration by parts. Let's set: - \( u = \sin^{n-1} x \) - \( dv = \sin x \, dx \) Then, we need to compute \( du \) and \( v \): 1. Differentiate \( u \): \[ du = (n-1) \sin^{n-2} x \cos x \, dx \] 2. Integrate \( dv \): \[ v = -\cos x \] Now, applying integration by parts, we have: \[ \int u \, dv = uv - \int v \, du. \] Substituting our expressions for \( u \), \( v \), \( du \), and \( dv \): \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = \int_{0}^{\frac{\pi}{2}} \sin^{n-1} x \sin x \, dx = \left[-\sin^{n-1} x \cos x \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x (n-1) \sin^{n-2} x \cos x \, dx. \] Evaluating the boundary term: - At \( x = 0 \): \( \sin^{n-1}(0) \cos(0) = 0 \) - At \( x = \frac{\pi}{2} \): \( \sin^{n-1}\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) = 0 \) Thus, the boundary term evaluates to zero: \[ \left[-\sin^{n-1} x \cos x \right]_{0}^{\frac{\pi}{2}} = 0. \] Now we focus on the integral: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \cos^2 x \, dx. \] Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the integral: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \left( \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx - \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx \right). \] Let \( I_n = \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx \). Then we have: \[ I_n = (n-1) \left( I_{n-2} - I_n \right). \] Rearranging gives: \[ I_n + (n-1) I_n = (n-1) I_{n-2}, \] which simplifies to: \[ n I_n = (n-1) I_{n-2}. \] Dividing both sides by \( n \): \[ I_n = \frac{n-1}{n} I_{n-2}. \] Thus, we have shown that: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx. \] This completes the proof of the reduction formula.