# Substitution Problem

• May 1st 2012, 09:58 AM
njdurkin
Substitution Problem
pg 339 problem # 25

Find the integral:

Attachment 23756

First of all, I don't even know what the sin to the 6 means. Does that mean take the sin of theta and then take that answer to the 6th power?
Secondly, I don't know where to start with this substitution problem. That exponent is throwing me off big time.
• May 1st 2012, 10:13 AM
TheIntegrator
Re: Substitution Problem
make the substitution $\displaystyle u= sin(\theta) => du= cos(\theta) d\theta$

Now our integral becomes $\displaystyle \int u^6 du$ which is much easier to solve => solve using the general power rule

Can you finish it off ?
• May 1st 2012, 10:24 AM
Prove It
Re: Substitution Problem
Quote:

Originally Posted by njdurkin
pg 339 problem # 25

Find the integral:

Attachment 23756

First of all, I don't even know what the sin to the 6 means. Does that mean take the sin of theta and then take that answer to the 6th power?
Secondly, I don't know where to start with this substitution problem. That exponent is throwing me off big time.

Yes, as you thought, \displaystyle \displaystyle \begin{align*} \sin^6{\theta} = \left(\sin{\theta}\right)^6 \end{align*} :)
• May 1st 2012, 11:07 AM
njdurkin
Re: Substitution Problem
Ah okay now it is easy. if sin to the six theta is simply sin theta to the power of six, then you can substitute 'just the sin theta' . I didn't know if that six in there made it one whole thing but now I see that it's simple. I've never seen that before. Yes I've got it! Thank you both! (:

There is a problem after that (#27) that is almost exactly the same thing:

Attachment 23757

I would expect everything to be the same except it is now 5 theta. But they are saying you should get (1/35) times sin of 5 theta to the seven instead of what I got which was (1/7) times that. Confused...
• May 1st 2012, 11:49 AM
TheIntegrator
Re: Substitution Problem
this problem is a little trickier => make the substitution $\displaystyle u=5\theta => du=5d\theta$ initially

now our integral becomes $\displaystyle \frac{1}{5} \int sin^{6}(u) cos(u) du$

now make the substitution $\displaystyle v=sin(u) => dv=cos(u) du$

$\displaystyle = \frac{1}{5} \int v^{6} dv$ now solve this simply integral using the general power rule.
• May 3rd 2012, 12:12 PM
njdurkin
Re: Substitution Problem
Awesome! Thanks.