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    How would you solve this Integration question?
    I used parts, but I had to do parts twice to it and then make it equal to I (which is some grade A work we did in class.)

    However is there an easier way to do it.

    The question is: The limits are pi/2 and 0.

    { 27sin2t.sint .dt =
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    (Original post by Nylex)
    Well, since sin 2t.sin 2t = sin^2 2t, you can use the double angle formula for cos 2t: cos 2t = 1 - 2sin^2 t => sin^2 2t = 0.5(1 - cos 4t).

    sorry I wrote it wrong, but Ive corrected it now.

    It was actually 27sin 2t. sint
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    (Original post by BlueAngel)
    How would you solve this Integration question?
    I used parts, but I had to do parts twice to it and then make it equal to I (which is some grade A work we did in class.)

    However is there an easier way to do it.

    The question is: The limits are pi/2 and 0.

    { 27sin2t.sint .dt =
    Use sin 2t = 2sin tcos t, so you have the integral of 27(2sin^2 tcos t) dt. Then you can use the substitution u = sin t, du = cos t dt.

    INT 27(2sin^2 tcos t) dt

    = 54 INT sin^2 tcos t dt

    u = sin t, du = cos t dt:

    = 54 INT u^2 du

    = 54.(1/3)u^3

    = 18sin^3 t

    Putting in the limits:

    = 18[(sin pi/2)^3 - (sin 0)^3]

    = 18
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    (Original post by BlueAngel)
    sorry I wrote it wrong, but Ive corrected it now.

    It was actually 27sin 2t. sint
    Yeah, no prob. Just seen that. See my post above.
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    (Original post by BlueAngel)
    sorry I wrote it wrong, but Ive corrected it now.

    It was actually 27sin 2t. sint

    Cant you just use substitution straightaway, I mean

    {27sin2t.sint.dt let U = sint so du/dt = cost and dt = 1/cost

    27 { sin2t. U . 1/cost

    27 { 2sintcost . U. 1/cost

    27 { 2U.U = 27 { 2U^2 . du


    Is that right????
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    (Original post by BlueAngel)
    Cant you just use substitution straightaway, I mean

    {27sin2t.sint.dt let U = sint so du/dt = cost and dt = 1/cost

    27 { sin2t. U . 1/cost

    27 { 2sintcost . U. 1/cost

    27 { 2U.U = 27 { 2U^2 . du


    Is that right????
    I'm not following that.. how are you going from du/dt = cos t to dt = 1/cos t? Full worked answer above btw.
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    (Original post by Nylex)
    I'm not following that.. how are you going from du/dt = cos t to dt = 1/cos t? Full worked answer above btw.


    Yes, I just used this method and it worked I got 18.

    If du/dt = cost you want to replace dt dont you.

    So you bring the du to the other side, so dt = du/cost. Which also is 1/cost .du.

    Get it?
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    you can actually spot that since sin t differentiates to cos t, so the integral of sin^2 t cos t will go exactly to some multiple of sin^3 t because of the chain rule. it's an exact integral.
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    (Original post by BlueAngel)
    Yes, I just used this method and it worked I got 18.

    If du/dt = cost you want to replace dt dont you.

    So you bring the du to the other side, so dt = du/cost. Which also is 1/cost .du.

    Get it?


    This is how I did it, wow Its easier than the teachers method.

    {27sin2t.sint.dt
    let U = sint, so du/dt = cost, and dt = 1/cost

    27 { sin2t. U . 1/cost

    27 { 2sintcost . U. 1/cost

    27 { 2U.U = 27 { 2U^2 . du

    Integrate 2U^2 = 2U^3/3.27.

    Replace U with sint. So you get 2sint^3/3

    Put the limits into it, which are pi/2 and 0.

    And you get 18 units squared.
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    (Original post by 4Ed)
    you can actually spot that since sin t differentiates to cos t, so the integral of sin^2 t cos t will go exactly to some multiple of sin^3 t because of the chain rule. it's an exact integral.

    Yeah youre right, thanx for pointing that out.
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    (Original post by 4Ed)
    you can actually spot that since sin t differentiates to cos t, so the integral of sin^2 t cos t will go exactly to some multiple of sin^3 t because of the chain rule. it's an exact integral.
    Oh crap, yeah. Inverse chain rule. Damn it.
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    PHEW! I dont know why you guys are all worried? You all seem to know everything about everything!!

    Im just gonna try lol .. off to do sum review questions
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    How about using the product-to-sum identity?
 
 
 
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