How to find this integral? guidance please.

Hi.

I need to find the following integral:

$\displaystyle \int \sin^6x\cos^2xdx$

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.

Thanks in advanced!

Re: How to find this integral? guidance please.

Quote:

Originally Posted by

**Stormey** I need to find the following integral:

$\displaystyle \int \sin^6x\cos^2xdx$

Have a look at this.

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Re: How to find this integral? guidance please.

Re: How to find this integral? guidance please.

Thank you.

that was very helpful!

Re: How to find this integral? guidance please.

OK, this is not working out.

this is what I've tried:

$\displaystyle \int \sin^6x\cos^3xdx$

$\displaystyle t=\cos x$

so we get:

$\displaystyle \int \sin^6xt^3xdx$

and then:

$\displaystyle t^3=\cos^3x\Rightarrow t^6=\cos^6x\Rightarrow t^6=(\cos^2x)^3\Rightarrow t^6=(1-\sin^2x)^3$

now what?

Re: How to find this integral? guidance please.

Hang on, is your integral $\displaystyle \displaystyle \begin{align*} \int{\sin^6{(x)}\cos^2{(x)}\,dx} \end{align*}$ like in the original post, or $\displaystyle \displaystyle \begin{align*} \int{\sin^6{(x)}\cos^3{(x)}\,dx} \end{align*}$ like in this latest post?

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Re: How to find this integral? guidance please.

Re: How to find this integral? guidance please.

OOPS! sorry...

it's:

$\displaystyle \int \sin^6x\cos^2xdx$

and then:

$\displaystyle t=\sin x\Rightarrow t^2=\sin^2x\Rightarrow t^2=1-\cos^2x\Rightarrow \cos^2x=1-t^2$

so:

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

how do I get rid of that dx?

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