Question:

<div><p>$$I_n = \int_{0}^{\pi/2} \sin^n x \, dx$$</p><p>$$I_n = \left[-\sin^{n-1} x \cos x\right]_0^{\pi/2} + \int_{0}^{\pi/2} (n-1) \sin^{n-2} x \cdot \cos^2 x \, dx = (n-1) \int_{0}^{\pi/2} \sin^{n-2} x \cdot (1 - \sin^2 x) \, dx$$</p><p>$$= (n-1) \int_{0}^{\pi/2} \sin^{n-2} x \, dx - (n-1) \int_{0}^{\pi/2} \sin^n x \, dx \Rightarrow I_n + (n-1)I_n = (n-1)I_{n-2}$$</p><p>$$I_n = \left(\frac{n-1}{n}\right) I_{n-2}$$</p><p><strong>Note:</strong></p><ol><li>$$\int_{0}^{\pi/2} \sin^n x \, dx = \int_{0}^{\pi/2} \cos^n x \, dx$$</li><li>$$I_n = \left(\frac{n-1}{n}\right)\left(\frac{n-3}{n-2}\right)\left(\frac{n-5}{n-4}\right) \dots I_0 \text{ or } I_1 \text{ according as } n \text{ is even or odd. } I_0 = \frac{\pi}{2}, I_1 = 1$$</li></ol><p>Hence $$I_n = \begin{cases} \left(\frac{n-1}{n}\right)\left(\frac{n-3}{n-2}\right)\left(\frac{n-5}{n-4}\right) \dots \left(\frac{1}{2}\right) \cdot \frac{\pi}{2}, & \text{if } n \text{ is even} \\ \left(\frac{n-1}{n}\right)\left(\frac{n-3}{n-2}\right)\left(\frac{n-5}{n-4}\right) \dots \left(\frac{2}{3}\right) \cdot 1, & \text{if } n \text{ is odd} \end{cases}$$</p></div>