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