need some explanation on integration

**Jones** · February 27th 2010, 10:35 AM

Hello,

I'm trying to integrate the following:

$\int sin^4 x\cdot cos^5 x$

rewrite:
$sin^4x(cosx\cdot cos^4 x)$

$cos^2 x = (1-sin^2x)$

$cos^4 x = (1-sin^2x)^2 = 1-2sin^2x+sin^4x$

multiply together:

$sin^4x-2sin^6x+sin^8x \cdot cosx$

let u = sin x, du = cos x

we have $u^4-2u^6+u^8 du$

My question is, the derivative of u is cos x

But we have 3 u terms. Aren't you getting a ´du´ for every u term?

**Ted** · February 27th 2010, 11:05 AM

Originally Posted by Jones

Hello,

I'm trying to integrate the following:

$\int sin^4 x\cdot cos^5 x$

rewrite:
$sin^4x(cosx\cdot cos^4 x)$

$cos^2 x = (1-sin^2x)$

$cos^4 x = (1-sin^2x)^2 = 1-2sin^2x+sin^4x$

multiply together:

$sin^4x-2sin^6x+sin^8x \cdot cosx$

let u = sin x, du = cos x

we have $u^4-2u^6+u^8 du$

My question is, the derivative of u is cos x

But we have 3 u terms. Aren't you getting a ´du´ for every u term?

Nope. only one $du$ , here:

Originally Posted by Jones

$sin^4x-2sin^6x+sin^8x \cdot cosx$
3

You should have brackets here.

It should be ${\color{red} ( } sin^4(x)-2sin^6(x)+sin^8(x) {\color{red} ) } cos(x)$

take $u=sin(x)$ to get:

$\int (u^4-2u^6+u^8)du$ which is easy.

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