# How to find this integral? guidance please.

• April 24th 2013, 10:49 AM
Stormey
How to find this integral? guidance please.
Hi.

I need to find the following integral:
$\int \sin^6x\cos^2xdx$
I just need a lead here.
How should I approach this?
I tried to use some trigonometric identities but couldn't get anywhere with it.
If it indeed can be simplified with trigonometric identities, please point me to the right direction.
• April 24th 2013, 10:58 AM
Plato
Re: How to find this integral? guidance please.
Quote:

Originally Posted by Stormey
I need to find the following integral:
$\int \sin^6x\cos^2xdx$

Have a look at this.
• April 24th 2013, 11:23 AM
ibdutt
Re: How to find this integral? guidance please.
• April 24th 2013, 11:43 AM
Stormey
Re: How to find this integral? guidance please.
Thank you.
• April 24th 2013, 12:30 PM
Stormey
Re: How to find this integral? guidance please.
OK, this is not working out.
this is what I've tried:

$\int \sin^6x\cos^3xdx$
$t=\cos x$
so we get:
$\int \sin^6xt^3xdx$
and then:
$t^3=\cos^3x\Rightarrow t^6=\cos^6x\Rightarrow t^6=(\cos^2x)^3\Rightarrow t^6=(1-\sin^2x)^3$
now what?
• April 24th 2013, 07:12 PM
Prove It
Re: How to find this integral? guidance please.
Hang on, is your integral \displaystyle \begin{align*} \int{\sin^6{(x)}\cos^2{(x)}\,dx} \end{align*} like in the original post, or \displaystyle \begin{align*} \int{\sin^6{(x)}\cos^3{(x)}\,dx} \end{align*} like in this latest post?
• April 24th 2013, 09:10 PM
ibdutt
Re: How to find this integral? guidance please.
• April 25th 2013, 12:32 AM
Stormey
Re: How to find this integral? guidance please.
OOPS! sorry...
it's:
$\int \sin^6x\cos^2xdx$

and then:
$t=\sin x\Rightarrow t^2=\sin^2x\Rightarrow t^2=1-\cos^2x\Rightarrow \cos^2x=1-t^2$

so:
$\int t^6(1-t^2)dx$

how do I get rid of that dx?
• April 25th 2013, 03:55 AM
ibdutt
Re: How to find this integral? guidance please.
If you put t = sin x then it is cosx dx = dt
Use the reduction formula Posted earlier