# Math Help - need some explanation on integration

1. ## need some explanation on integration

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?

2. 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.