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 \( dv \): \[ \int \sin^{n} x \, dx = \int \sin^{n-1} x \sin x \, dx = \left[-\sin^{n-1} x \cos x \right]_{0}^{\frac{\pi}{2}} + \int \cos x \cdot (n-1) \sin^{n-2} x \cos x \, dx. \] 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 0, and we are left with: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \cos^2 x \, dx. \] Now, we can express \( \cos^2 x \) in terms of \( \sin^{n-2} x \): Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can rewrite the integral: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x (1 - \sin^2 x) \, dx. \] This gives us: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \left( \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx - \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx \right). \] Rearranging this, we have: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx + (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx. \] This simplifies to: \[ n \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = (n-1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx. \] Dividing both sides by \( n \): \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx. \] Thus, we have shown the desired reduction formula: \[ \int_{0}^{\frac{\pi}{2}} \sin^{n} x \, dx = \frac{n-1}{n} \int_{0}^{\frac{\pi}{2}} \sin^{n-2} x \, dx. \]