# Integral powers of sine and cosine

• Mar 8th 2010, 11:26 AM
mj.alawami
Integral powers of sine and cosine
Question :

$\displaystyle \int sin^4.cos^2x.dx$

Attempt :
I reached a step where >>

$\displaystyle \frac{1}{8} \int (1-2cos2x+cos^2x).(1+cos2x)$

How to expand this ?
• Mar 8th 2010, 11:28 AM
Miss
Quote:

Originally Posted by mj.alawami
Question :

$\displaystyle \int sin^4.cos^2x.dx$

Attempt :
I reached a step where >>

$\displaystyle \frac{1}{8} \int (1-2cos2x+cos^2x).(1+cos2x)$

How to expand this ?

By the same method of expanding $\displaystyle (a+b+c)(d+e)$.
• Mar 8th 2010, 12:35 PM
Quote:

Originally Posted by mj.alawami
Question :

$\displaystyle \int sin^4.cos^2x.dx$

Attempt :
I reached a step where >>

$\displaystyle \frac{1}{8} \int (1-2cos2x+cos^2x).(1+cos2x)$

How to expand this ?

$\displaystyle \int{sin^4xcos^2x}dx=\int{sin^2xsin^2xcos^2x}dx$

$\displaystyle =\int{\left(1-cos^2x\right)\left(1-cos^2x\right)cos^2x}dx$

$\displaystyle =\int{\left(1-2cos^2x+cos^4x\right)cos^2x}dx$

may be more straightforward.

Also you have erred on the way...

$\displaystyle sin^2xsin^2x=\frac{1}{4}(1-cos2x)(1-cos2x)=\frac{1}{4}\left(1-2cos2x+cos^2(2x)\right)$
• Mar 8th 2010, 12:49 PM
wonderboy1953
An alternate way
You can try integration by parts on the following:

$\displaystyle \int(sin^4xcosx)cosxdx$

letting $\displaystyle u = cosx$ and $\displaystyle dv = sin^4xcosx dx$