Integrate

**Morgan82** · December 6th 2009, 03:08 PM

(cosx)^3 * (sinx)^2 dx

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

**VonNemo19** · December 6th 2009, 03:17 PM

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

**Prove It** · December 6th 2009, 03:32 PM

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 $u = \sin{x}$ .

**VonNemo19** · December 6th 2009, 03:36 PM

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.

**Prove It** · December 6th 2009, 03:48 PM

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.

**VonNemo19** · December 6th 2009, 03:50 PM

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.

**Morgan82** · December 6th 2009, 04:03 PM

I got it. Thanks

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