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Thread: integration part 2

  1. #1
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    integration part 2

    $\displaystyle
    \int (xsinx^2cosx^2),dx

    $

    all i need are steps leading to the answer
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  2. #2
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    Start by makin' $\displaystyle u=x^2.$
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  3. #3
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    Red face integration part 2

    ok not quite getting it. can you help me out a little more please.
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  4. #4
    Math Engineering Student
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    $\displaystyle u=x^2\implies du=2x\,dx,$

    $\displaystyle \int {x\sin x^2 \cos x^2 \,dx} = \frac{1}
    {2}\int {\sin u\cos u\,du} = \frac{1}
    {4}\int {\sin 2u\,du} .$
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  5. #5
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    needing a little more help w/ this....

    ok so the answer is

    $\displaystyle

    (x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C
    $

    and i'm not sure how i would get that using

    $\displaystyle
    {1/4} \int sin2u

    $
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by simsima_1 View Post
    ok so the answer is

    $\displaystyle

    (x)tan^{-1}(x)-\frac{1}{2}ln({x^2}+1)+C
    $

    and i'm not sure how i would get that using

    $\displaystyle
    {1/4} \int sin2u

    $
    there are often many different forms for the answer to an integral. following Krizalid's suggestion is good enough.
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