Proving a reduction formula for the antiderivative of $\cos^n(x)$ 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. 

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.
