# Derivation and application of reduction formula?

## "Use integration by parts to derive the reduction formula $\setminus \int \setminus {\cos}^{n} \left(x\right) \mathrm{dx} = \frac{1}{n} \setminus \sin x \setminus {\cos}^{n - 1} \left(x\right) + \frac{n - 1}{n} \setminus \int \setminus {\cos}^{n - 2} \left(x\right) \mathrm{dx}$, where $n$ is a positive integer." I assume that I need to split this into $\setminus \int \setminus \cos x \setminus {\cos}^{n - 1} x$ or something to get a valid IBP process? Use the previous reduction formula or integration by parts to evaluate: $\setminus \int \setminus {\cos}^{3} \mathrm{dx}$ (Where is the $x$ in this problem??)

Apr 9, 2018

Write the integrand as:

${\cos}^{n} x = {\cos}^{n - 1} x \cos x = {\cos}^{n - 1} x d \left(\sin x\right)$

then we can integrate by parts:

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

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

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

Now use: ${\sin}^{2} x = 1 - {\cos}^{2} x$:

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

and using the linearity of the integral:

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

The integral:

${I}_{n} = \int {\cos}^{n} x \mathrm{dx}$

appears now on both sides of the equation and we can solve for it:

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

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

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

which proves the reduction formula.

For $n = 3$ in particular we have:

$\int {\cos}^{3} x \mathrm{dx} = \frac{{\cos}^{2} x \sin x}{3} + \frac{2}{3} \int \cos x \mathrm{dx}$

$\int {\cos}^{3} x \mathrm{dx} = \frac{{\cos}^{2} x \sin x + 2 \sin x}{3} + C$

and simplifying:

$\int {\cos}^{3} x \mathrm{dx} = \frac{\left(1 - {\sin}^{2} x\right) \sin x + 2 \sin x}{3} + C$

$\int {\cos}^{3} x \mathrm{dx} = \frac{3 \sin x - {\sin}^{3} x}{3} + C$

$\int {\cos}^{3} x \mathrm{dx} = \sin x - {\sin}^{3} \frac{x}{3} + C$