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Math Help - Substitution Problem

  1. #1
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    Substitution Problem

    pg 339 problem # 25

    Find the integral:

    Substitution Problem-adam.gif

    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.
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  2. #2
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    Re: Substitution Problem

    make the substitution u= sin(\theta) => du= cos(\theta) d\theta

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

    Can you finish it off ?
    Last edited by TheIntegrator; May 1st 2012 at 10:16 AM.
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  3. #3
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    Re: Substitution Problem

    Quote Originally Posted by njdurkin View Post
    pg 339 problem # 25

    Find the integral:

    Click image for larger version. 

Name:	adam.gif 
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ID:	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 \begin{align*} \sin^6{\theta} = \left(\sin{\theta}\right)^6 \end{align*}
    Thanks from njdurkin
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  4. #4
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    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:

    Substitution Problem-what.gif

    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...
    Last edited by njdurkin; May 1st 2012 at 11:11 AM.
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  5. #5
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    Re: Substitution Problem

    this problem is a little trickier => make the substitution u=5\theta => du=5d\theta initially

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

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

    = \frac{1}{5} \int v^{6} dv now solve this simply integral using the general power rule.
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  6. #6
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    Re: Substitution Problem

    Awesome! Thanks.
    Last edited by njdurkin; May 3rd 2012 at 12:39 PM.
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