# Thread: [SOLVED] Integral of cos squared x

1. ## [SOLVED] Integral of cos squared x

$\displaystyle \int sin^{2}xcosx dx$

so using
$\displaystyle sin^{2}x = 1 - cos^{2}x$
$\displaystyle \int sin^{2}xcosx dx = \int cosx - cos^{3}x dx$

which is all fine and dandy (i think) but what's the integral of
$\displaystyle cos^{3}x$?

is there a better way of doing it?

Also what would be
$\displaystyle \int sin^{5}x dx$or $\displaystyle \int cos^{5}x dx$ and $\displaystyle \int cos^{7}x dx$

i guess im looking for a general rule of
$\displaystyle \int cos^{n}x dx$ and $\displaystyle \int sin^{n}x dx$

2. Originally Posted by UbikPkd
$\displaystyle \int sin^{2}xcosx dx$

so using
$\displaystyle sin^{2}x = 1 - cos^{2}x$
$\displaystyle \int sin^{2}xcosx dx = \int cosx - cos^{3}x dx$

which is all fine and dandy (i think) but what's the integral of
$\displaystyle cos^{3}x$?

INT.[sin^2(X)cosX]dX
= INT.[sin^2(X)]cosX dX

if u were sinX, then du = cosX dX, so

= (1/3)sin^3(X) +C

3. I personally would use u-substitution. Notice that $\displaystyle \cos{x}$ is the derivative of $\displaystyle \sin{x}$, so it works nicely.

Let $\displaystyle u=\sin{x}$, then $\displaystyle \,du=\cos{x}\,dx$

So that turns the integral into

$\displaystyle \int u^2 \,du=\frac{u^3}{3}+C$

Now substituting $\displaystyle \sin{x}$ back into u:

$\displaystyle =\frac{(\sin{x})^3}{3}+C$

(Didn't see u post ticbol) :P

4. thats great, thanks a lot!

5. Unfortunately there is no easy generalisation for nth powers, you have to use reduction formulas. Here are some of them. You could try to derive them using integration by parts, if you've learnt that.

6. i had a feeling it wasn't gonna be simple....thanks anyway!