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I want to show that for all $n\ge 2$, it holds that $$ \int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx. $$ I'm not even getting the result for the induction base $(n=2)$: Using integration by parts, I only get $$ \int \cos^2 x\ dx = \cos x \sin x + \int \sin^2 x\ dx. $$ I'm suspecting that I need to use some trigonometric identity here.

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Possible duplicate: math.stackexchange.com/questions/193435/… –  Joe Nov 13 '12 at 17:40
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2 Answers

up vote 2 down vote accepted

You only need that $$\sin^2 x+\cos^2 x=1$$

Let $$\varphi(n)=\int \cos^n x dx$$

Integrate by parts

$$\begin{cases}\cos^{n-1} x =u\\ \cos x dx =dv\end{cases}$$

Then

$$\begin{align}\varphi(n)&=\int \cos^n x dx\\ &=uv-\int v du\\&=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x \sin ^2x d x\end{align}$$

But $\sin^2 x=1-\cos^x $

$$\begin{align}&=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x \sin ^2x d x \\&=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x \left(1-\cos^2 x\right) d x \\ &=\cos^{n-1}x\sin x+\int (n-1)\cos^{n-2}x dx-\int (n-1)\cos^{n-2}x \cos^2 x d x \\&=\cos^{n-1}x\sin xdx+(n-1)\int \cos^{n-2}x dx-(n-1)\int \cos^{n }x dx \\&=\cos^{n-1}x\sin xdx+(n-1)\int \cos^{n-2}xdx -(n-1)\varphi(n)\end{align}$$

Now solve for $\varphi(n)$.

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thanks, great answer! :) –  somebody Nov 13 '12 at 18:23
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For the base case, use that $\sin^2 x = 1-\cos^2 x$, write $1=\cos^{0} x$ and you get:

$$\int \cos^2 x \ dx = \cos x \sin x + \int \cos^{0} x\ dx- \int \cos^2 x \ dx$$

So: $$2\int \cos^2 x dx = \sin x \cos x + \int \cos^0 x \ dx$$

Dividing by $2$ gives your result.

Now try for $n>2$. It's basically the same argument, where you replace $\sin^2 x=1-\cos^2 x$ and then solve for the integral.

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