# Integration Help on trigo

• Dec 14th 2012, 03:56 AM
tempq1
Integration Help on trigo
$\displaystyle \int cos^4 2x sin^3 2x dx$
I've tried simplifying the equation first but got stucked instead, and tried substitution also.
Which method should i use for this?
• Dec 14th 2012, 04:21 AM
kspkido
Re: Integration Help on trigo
hmmm why don't you try to first factor sin^3(2x) to sin^2(2x)(sin(2x))... then since sin^2(2x) is also equal to 1-cos^2(2x).... you rearrange it to a indefinite integral of cos^4(2x)[1-cos^2(2x)]sin(2x)dx.. :) then do substitution... u=cos (2x).... :) du= (2)(-sin(2x))dx.... now... please do the rest :)
• Dec 14th 2012, 04:23 AM
MarkFL
Re: Integration Help on trigo
Try rewriting the integrand as follows:

$\displaystyle \cos^4(2x)\sin^3(2x)=\cos^4(2x)\sin^2(2x)\sin(2x)= \cos^4(2x)(1-\cos^2(2x))\sin(2x)=$

$\displaystyle \cos^4(2x)\sin(2x)-\cos^6(2x)\sin(2x)$

Can you proceed from here using an appropriate substitution?
• Dec 14th 2012, 08:50 AM
hollywood
Re: Integration Help on trigo
At my university, students are taught general methods for the integrals $\displaystyle \int\cos^nx\sin^mx\ dx$ and $\displaystyle \int\sec^nx\tan^mx\ dx$. The substitution depends on whether n and m are even or odd - it's actually pretty straightforward if you look at the derivatives of those four trigonometric functions.

- Hollywood
• Dec 14th 2012, 06:29 PM
tempq1
Re: Integration Help on trigo
Thanks guys question solved. Just another question
$\displaystyle \int cot^2 \pi x dx$
i tried changing it into
$\displaystyle \frac{1 + cos2\pi x}{1-cos2\pi x}$ and got stuck.
ps:can't seem to edit my original post.
• Dec 14th 2012, 10:36 PM
MarkFL
Re: Integration Help on trigo
I would use a Pythagorean identity to rewrite the integrand as:

$\displaystyle \csc^2(\pi x)-1$

and then use the fact that:

$\displaystyle \frac{d}{dx}(\cot(u(x)))=-\csc^2(u(x))\frac{du}{dx}$

to rewrite the integral as:

$\displaystyle -\frac{1}{\pi}\int\,d(\cot(\pi x))-\int\,dx$