# Integrate

• Dec 6th 2009, 03:08 PM
Morgan82
Integrate
(cosx)^3 * (sinx)^2 dx

Can someone tell me how to start this, how to get at it? Thanks
• Dec 6th 2009, 03:17 PM
VonNemo19
Quote:

Originally Posted by Morgan82
(cosx)^3 * (sinx)^2 dx

Can someone tell me how to start this, how to get at it? Thanks

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

Let $u=\sin{x}$...
• Dec 6th 2009, 03:32 PM
Prove It
Quote:

Originally Posted by VonNemo19
$=\int(1-\sin^2x)\sin^2x\cos{x}dx$

Let $u=\sin{x}$...

$= \int{(\sin^2{x} - \sin^4{x})\cos{x}\,dx}$.

NOW let $u = \sin{x}$.
• Dec 6th 2009, 03:36 PM
VonNemo19
Quote:

Originally Posted by Prove It
$= \int{(\sin^2{x} - \sin^4{x})\cos{x}\,dx}$.

NOW let $u = \sin{x}$.

I don't see anything wrong with

$\int(1-u^2)u^2du=\int(u^2-u^4)du$

The distribution can come after or before.
• Dec 6th 2009, 03:48 PM
Prove It
True, there's nothing wrong with it.

It's just that if a student is trying to get a function to be of the form $f\left(u(x)\right)\cdot\frac{du}{dx}$, it's hard enough to find the correct product as it is, let alone having a second product in there.
• Dec 6th 2009, 03:50 PM
VonNemo19
Quote:

Originally Posted by Prove It
True, there's nothing wrong with it.

It's just that if a student is trying to get a function to be of the form $f\left(u(x)\right)\cdot\frac{du}{dx}$, it's hard enough to find the correct product as it is, let alone having a second product in there.

Fair enough.
• Dec 6th 2009, 04:03 PM
Morgan82
I got it. Thanks :)