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    how would you integrate sin^2tcost??
    I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?
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    The one that jumps out at me is integration by parts but it'd be a real pig.

    Practice questions and you should start spotting them.
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    I'm not too sure what you mean by spotting the pattern. Reverse chain rule?

    In regards to sin^2(t)cos(t) try a substitution of u=sin(t)
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    (Original post by hajs)
    how would you integrate sin^2tcost??
    I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?
    It is obvious from the inverse chain rule

    If you want ... You can substitute u=sint ... However, spotting that is the same as seeing the inverse chain rule
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    (Original post by hajs)
    how would you integrate sin^2tcost??
    I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?
    You need to use a substitution u, so that \frac{du}{dt} is equal to whatever is multiplied by the function of u.

    In this case, the substitution is u = \sin t because when differentiated, this become \cos t, which cancels with the \cos t in the integrand.
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    (Original post by SeanFM)
    The one that jumps out at me is integration by parts but it'd be a real pig.

    Practice questions and you should start spotting them.
    Yeah i was thinking that.. not worth it right?
    And mmm yeah but i don't understand the method behind it initially.
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    (Original post by morgan8002)
    You need to use a substitution u, so that \frac{du}{dt} is equal to whatever is multiplied by the function of u.

    In this case, the substitution is u = \sin t because when differentiated, this become \cos t, which cancels with the \cos t in the integrand.
    how did you know that u=sint.. this would have taken me a while to figure out.. guess practice is the key???
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    (Original post by SeanFM)
    The one that jumps out at me is integration by parts but it'd be a real pig.

    Practice questions and you should start spotting them.
    IBP is actually very easy with this question
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    (Original post by hajs)
    how did you know that u=sint.. this would have taken me a while to figure out.. guess practice is the key???
    As I said ... That understanding is identical to the understanding needed to use inverse chain rule
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    (Original post by hajs)
    how did you know that u=sint.. this would have taken me a while to figure out.. guess practice is the key???
    Just because \frac{d(\sin t)}{dt} = \cos t.
    You need to find a substitution that cancels the other term (that cannot be put in terms of the substitution).


    After practice, you will be able to use inverse chain rule, which is quicker and based on intuition.
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    (Original post by TenOfThem)
    As I said ... That understanding is identical to the understanding needed to use inverse chain rule
    (Original post by morgan8002)
    Just because \frac{d(\sin t)}{dt} = \cos t.
    You need to find a substitution that cancels the other term (that cannot be put in terms of the substitution).


    After practice, you will be able to use inverse chain rule, which is quicker and based on intuition.
    okay thank you both!! I will practice more of these
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    (Original post by hajs)
    okay thank you both!! I will practice more of these
    You're welcome.
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    s(sin^2tcost) dx
    u= (sint)^2
    du/dx= 2sintcost
    dv/dx=cost
    v=sint therefore s(sin^2tcost) dx= sint^3t-[s2sintcos^2tdx] apply again and it should come out
    ie u=cos^2t
    du/dx= -2sint
    dv/dx= 2sint
    v= -2cost
    therefore, sin^3t + 2cos^3t -[s 2sintcostdx]
    where -[s2sintcostdx]=
    u=2sint
    du/dx= 2cost
    dv/dx=cost
    v=sint keep applying until sin and cos int or you get =y where you sub.

    or you can sub with u so that you get rid of a 2cost
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    (Original post by noumenon)
    s(sin^2tcost) dx
    u= (sint)^2
    ....
    Really, horrendously overcomplicated!

    The substitution u = sin t is all that's required here
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    (Original post by davros)
    Really, horrendously overcomplicated!

    The substitution u = sin t is all that's required here
    I've always preferred by parts than substitution There was nothing horrendous about it at all!
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    (Original post by hajs)
    how would you integrate sin^2tcost??
    I was looking on exam solutions and he seemed to use the method where you easily can spot the pattern, i find this method hard and wonder if there is another way round it?
    let u = sin(t)
    du = cos(t)dt

    = ∫ u^2du
    ...u^3
    =------- + C
    ....3
 
 
 
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