# Show that  I_n = -1/n \ cosx \ sin^(n-1)x+(n-1)/n \ I_(n-2)  where I_n=int \ sin^nx \ dx , and n ge 2?

Sep 24, 2017

We want to show that:

${I}_{n} = - \frac{1}{n} \setminus \cos x \setminus {\sin}^{n - 1} x + \frac{n - 1}{n} \setminus {I}_{n - 2}$

Where ${I}_{n} = \int \setminus {\sin}^{n} x \setminus \mathrm{dx}$, and $n \ge 2$.

We can write the integral as:

${I}_{n} = \int \setminus \left(\sin x\right) \left({\sin}^{n - 1} x\right) \setminus \mathrm{dx}$ provided $n \ge 2$

We can use integration by parts:

Let $\left\{\begin{matrix}u & = {\sin}^{n - 1} x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = \left(n - 1\right) {\sin}^{n - 2} x \cos x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = \sin x & \implies v & = - \cos x\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us:

$\int \setminus \left({\sin}^{n - 1} x\right) \left(\sin x\right) \setminus \mathrm{dx} = \left({\sin}^{n - 1} x\right) \left(- \cos x\right) - \int \setminus \left(- \cos x\right) \left(\left(n - 1\right) {\sin}^{n - 2} x \cos x\right) \setminus \mathrm{dx}$

$\therefore {I}_{n} = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \int \setminus {\cos}^{2} x \setminus {\sin}^{n - 2} x \setminus \mathrm{dx}$

Using the identity, ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have:

${I}_{n} = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \int \setminus \left(1 - {\sin}^{2} x\right) \setminus {\sin}^{n - 2} x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \int \setminus {\sin}^{n - 2} x - {\sin}^{n} x \setminus \mathrm{dx}$
$\setminus \setminus \setminus = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) \left({I}_{n - 2} - {I}_{n}\right)$
$\setminus \setminus \setminus = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) {I}_{n - 2} - \left(n - 1\right) {I}_{n}$

${I}_{n} + \left(n - 1\right) {I}_{n} = - \cos s x \setminus {\sin}^{n - 1} x + \left(n - 1\right) {I}_{n - 2}$

$\therefore n {I}_{n} = - \cos x \setminus {\sin}^{n - 1} x + \left(n - 1\right) {I}_{n - 2}$

$\therefore {I}_{n} = - \frac{1}{n} \cos s x \setminus {\sin}^{n - 1} x + \frac{n - 1}{n} {I}_{n - 2}$ QED