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Math Help - Integration

  1. #1
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    Integration

    These few questions have been bugging me for hours, help would be appreciated
    Attached Thumbnails Attached Thumbnails Integration-integration.bmp  
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Integration

    1. What's the lower bound? Use the goniometric substitution t=\tan(u)
    2. Use integration by parts.
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  3. #3
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    Re: Integration

    Quote Originally Posted by Siron View Post
    1. What's the lower bound? Use the goniometric substitution t=\tan(u)
    2. Use integration by parts.
    Thank you How do you know when to use t = tanu?

    Also would you be able to give me an indication of what to do for this one:

    Integrate 9t/(9t^2 + 6t +5)
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Integration

    Quote Originally Posted by qwerty31 View Post
    Thank you How do you know when to use t = tanu?
    Because with that subsitution you can 'eliminate' the root form which is easier to work with. (Have you solved the integral?)

    Quote Originally Posted by qwerty31 View Post
    Also would you be able to give me an indication of what to do for this one:
    Integrate 9t/(9t^2 + 6t +5)
    \int \frac{9t}{9t^2+6t+5}dt = \frac{1}{2} \int \frac{(18t+6)-6}{9t^2+6t+5}dt = \frac{1}{2} \left(\int \frac{18t+6}{9t^2+6t+5}dt - \int \frac{6}{9t^2+6t+5}dt\right)

    I guess you can take it from here. To solve \int \frac{6}{9t^2+6t+5}dt write 9t^2+6t+5=(3t+1)^2+4 (because D<0)
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  5. #5
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    Re: Integration

    sin^2u+ cos^2u= 1 so, dividing both sides by cos^2 u, tan^2u+ 1= sec^2u. Because of that the substitution x= tan u changes \sqrt{1+ x^2} to \sqrt{1+ tan^2 u}= \sqrt{sec^2 u}= sec u. Generally speaking, any time you have a square root of a quadratic, you should consider a trig substitution. (Some people prefer hyperbolic functions which satisfy similar identities.)

    As for your new question, first complete the square: 9t^2+ 6t+ 5= 9(t^2+ (2/3)t+ 1/9- 1/9)+ 5)= 9(t+ 1/3)^2+ 4.
    Now the substitution x= t+ 1/3 changes to integrand to \frac{9u- 3}{9u^2+ 4}= \frac{u}{u^2+ (2/3)^2}+ (4/9)\frac{1}{(3u/2)^2+ 1}
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