# Integrating $\int \sin^n{x} \ dx$

I am working on trying to solve this problem:

Prove: $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$

Here are the steps that I follow in the example that I am reading:

$u = \sin^{n - 1}{x}$

$du = (n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx$

$v = -\cos{x}$

$dv = \sin{x} \ dx$

$\int \sin^n{x} \ dx = \sin^{n - 1}{x} \cdot \sin{x} \ dx$

$\int \sin^n{x} \ dx = \underbrace{\sin^{n - 1}{x}}_{u} \cdot \underbrace{-\cos{x}}_{v} - \int \underbrace{-\cos{x}}_{v} \cdot \underbrace{(n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx}_{du}$

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

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

Here is where I get lost. How did we go from $\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$ to $\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$? Even more specifically, where did $\sin^{n}{x}$ come from?

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

I get this part.

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

$\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$

Could someone please explain what I am missing?

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Looks like he just cleared the parentheses, then split the integral into 2. – Mike Feb 24 '12 at 0:59
$\sin^{n-2}x\sin^2x=\sin^nx$ – Gerry Myerson Feb 24 '12 at 1:06
ohhh... then the $(n - 1)$ was distributed when the integral was broken up, I guess that is what was throwing me off. Thanks, Mike! – Oliver Spryn Feb 24 '12 at 1:07
... and Gerry!! – Oliver Spryn Feb 24 '12 at 1:07
I think you might have an extra x in the first term of the answer (by mistake) See the answer below. – Kirthi Raman Feb 24 '12 at 3:00

First let us denote

$$I_n = \int \sin^n{x} \ dx$$

$$\int u(x) v'(x) dx = u(x) v(x) - \int v(x) u' (x) dx$$

Here $u(x) = \sin^{n-1}{x} \hspace{3pt}$ and $\hspace{3pt} v'(x) = \sin x$

$\Rightarrow v(x) = -\cos x$

Therefo‌‌‌‌‌‌‌‌re‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ \begin{align*} I_n &= -\cos x \hspace{3pt} \sin^{n-1}x + \int \cos^2 x \hspace{4pt} (n-1) \sin^{n-2} x \hspace{4pt} dx \\ &= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx - (n-1) \int \sin^{n} dx\\ &= -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x \hspace{4pt} dx - (n-1) I_n \end{align*}

$$\Rightarrow (1+n-1)I_n = -\cos x \hspace{3pt} \sin^{n-1}x + (n-1) \int \sin^{n-2} x dx$$

$$\Rightarrow I_n = \frac{-\cos x \hspace{3pt} \sin^{n-1}x}{n} + \frac{(n-1)}{n} \int \sin^{n-2} x dx$$

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