1. ## reduction formula proof

How would I prove the following reduction formula.

∫cos^n xdx=1/n (cos^n-1 x)sinx+(n-1/n) ∫cos^n-2 xdx?

I know there are plenty of sites that explain this but I want an answer from a trusted source.

2. Originally Posted by gammaman
How would I prove the following reduction formula.

∫cos^n xdx=1/n (cos^n-1 x)sinx+(n-1/n) ∫cos^n-2 xdx?

I know there are plenty of sites that explain this but I want an answer from a trusted source.
Hi

Let $I_n = \int \cos^nx \:dx$

$I_n = \int \cos^2x \:cos^{n-2}x \:dx$

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

$I_n = \int cos^{n-2}x \:dx - \int \sin^2x \:cos^{n-2}x \:dx$

$I_n = \int cos^{n-2}x \:dx - \int \sin x \:\sin x\:cos^{n-2}x \:dx$

By parts with $u = \sin x$ and $dv = \sin x\:cos^{n-2}x \:dx = -\frac{1}{n-1}\:d\left(cos^{n-1}x\right)$

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

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

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

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

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