\(I_n= \int\limits^\infty_0 e^{-x} sin^nx\,dx\)

\(= \int\limits^\infty_0\underset{I}{\underline{e^{-x}sin^{n-1}x}}\,\,\underset{II}{\underline {sinx\, dx}}\)

\(= \left[e^{-x}sin^{n -1}x\int sinx\,dx - \int\frac{e^{-x}(n - 1)sin^{n -2}x cos x - sin^{n -1}x}{x - cos x \,dx}\right]^\infty_0\)

\(= \left[-e^{-x}sin^{n -1}x cos\,x\right]^\infty _0 + \int\limits ^\infty_0(n - 1)e^{-x}sin^{n -2}x . cos^2x \, dx \\\,\,\,\,\,\,\,\,\, -\int\limits ^\infty_0e^{-x} sin^{n -1} x\, dx\)

\(= (n - 1)\int\limits^\infty_0e^{-x}sin^{n -2}x\,dx - \int\limits^\infty_0(n - 1)e^{-x}sin^nx\,dx -\int\limits ^\infty_0e^{-x} sin^{n -1} x\, cosx\,dx\)

(As \(e^{-\infty} = 0\) and \(cos^2x = 1 - sin^2x\))

\(= (n - 1)I_{n -2}- (n -1) I_n - \left[e^{-x}\frac{sin^nx}{n}\right]^\infty_0 +\int\limits^\infty_0- e^{-x} \frac{sin^nx}{n}dx\)

⇒ \(I_n = (n - 1)I_{n - 2} - (n - 1)I_n - \frac1nI_n\)     \((\because e^{-0} = 0)\)

⇒ \(I_n + (n - 1)I_n + \frac 1nI_n = (n - 1)I_{n -2}\)

⇒ \(I_n = \frac{n - 1}{1 + n - 1 + \frac1n}\) \(I_{n - 2} = \frac{n(n - 1)}{n^2 + 1}I_{n -2}\)

⇒ \(I_n = \frac{n(n - 1)}{n^2 + 1}I_{n - 2}\)

which is required relation.