Trigonometric Integral Homework: Solving \int (sin^6(x))(cos^3(x))

[Mugen Prospec]

Homework Statement

$$\int$$ 41(sin$$^{6}$$(x))(cos$$^{3}$$(x))

Homework Equations

The Attempt at a Solution

I think you are supposed to use the half angle identities and then maybe integration by parts but I am lost on it.

 

[lanedance]

how about starting by trying the substitution u = sinx

 

[Mugen Prospec]

I have tried a few different things, more than I wanted to list but it just keeps getting more and more convoluted. I need a walk through we just started this in class and my teacher doesn't answer questions so I am just a bit lost over all.

 

[Dick]

You don't have to do anything complicated with it, cos(x)dx=d(sin(x)). Just substitute u=sin(x). There are rules for dealing with powers of sin's and cos's. They are particularly easy if one power is odd.

 

[Mugen Prospec]

You don't have to do anything complicated with it, cos(x)dx=d(sin(x)). Just substitute u=sin(x). There are rules for dealing with powers of sin's and cos's. They are particularly easy if one power is odd.

Ok I know what your talking about. It was just in our chapter about integration by parts so I was a little first sight shocked. Can some one give me an answer so when I complete it I can know if I am correct or not.

 

[Dick]

Oh, come on. Just work it out and show us what you get. I'll guarantee someone will check it.

 

[Mugen Prospec]

Oh, come on. Just work it out and show us what you get. I'll guarantee someone will check it.

Ok I have other work to do this one has bee on my mind all night. Ill post it tomorrow when I am clear of thought

 

[Dick]

Ok I have other work to do this one has bee on my mind all night. Ill post it tomorrow when I am clear of thought

It's REALLY easy with the substitution lanedance suggested. You might want to clear your mind on this one and go to bed happy. But tomorrow is ok too.

 

[Mugen Prospec]

Ok thanks a lot Ill see what I can do tonight.

 

[Dick]

Free hint since you are playing along: cos^2(x)=1-sin^2(x)=(1-u^2).

 

[Mugen Prospec]

ok i got
41(sin^7(x)/7)(1/2 x + 1/4sin2x+c)

 

[lanedance]

doesn;t look quite right to me, maybe show your working

 

[Mugen Prospec]

Thats what I was thinking. after using u substitution I was left with

(u)^6 (cos^2(x)) cos(x) du/cos(x)

So cos(x) canceled out.

 

[Dick]

Thats what I was thinking. after using u substitution I was left with

(u)^6 (cos^2(x)) cos(x) du/cos(x)

So cos(x) canceled out.

Fine. Now what's cos^2(x) in terms of u?

 

[Mugen Prospec]

Oh ok the identity. so now all turns into
u^7 (1-u^2) du
Do I do their antiderivative now? And then substitute sin(x) back in. I am not sure if you can do each of there AD since they are multiplying one another.

 

[Dick]

Oh ok the identity. so now all turns into
u^7 (1-u^2) du
Do I do their antiderivative now? And then substitute sin(x) back in. I am not sure if you can do each of there AD since they are multiplying one another.

Of course you don't take the AD of each one. That's wrong. You multiply it out.

 

[Mugen Prospec]

OK that what I thought I am trying to do this not without witting it down since I am in chemistry.
So we get (u^6)-(u^8)du
then (u^7)/7) - (u^9)/9)
41 (sin^7(x))/7) - (sin^9(x))/9)
is that it maybe?

 

[Dick]

OK that what I thought I am trying to do this not without witting it down since I am in chemistry.
So we get (u^6)-(u^8)du
then (u^7)/7) - (u^9)/9)
41 (sin^7(x))/7) - (sin^9(x))/9)
is that it maybe?

You are missing a parenthesis level following the 41, but yes, that's it.

 

[Mugen Prospec]

Ok awesome thank you for you patience.