# [SOLVED] Integral of cos squared x

• Oct 12th 2007, 11:47 PM
UbikPkd
[SOLVED] Integral of cos squared x
$\int sin^{2}xcosx dx$

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

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

is there a better way of doing it?

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

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

• Oct 13th 2007, 12:01 AM
ticbol
Quote:

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

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

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

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
• Oct 13th 2007, 12:04 AM
DivideBy0
I personally would use u-substitution. Notice that $\cos{x}$ is the derivative of $\sin{x}$, so it works nicely.

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

So that turns the integral into

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

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

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

(Didn't see u post ticbol) :P
• Oct 13th 2007, 12:10 AM
UbikPkd
thats great, thanks a lot! :)
• Oct 13th 2007, 12:33 AM
DivideBy0
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.
• Oct 13th 2007, 12:42 AM
UbikPkd
i had a feeling it wasn't gonna be simple....thanks anyway!