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Math Help - How do I integrate this?

  1. #1
    Member sinewave85's Avatar
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    How do I integrate this?

    I am stuck!

    \int x^{3}\cos{\tan^{-1}{x}}

    I tried

    \mbox{Let } u=\tan^{-1}{x}, du=\frac{1}{x^{2} +1}

    Then with some substituting,

    \int x^{3}\cos{\tan^{-1}{x}} = \int\sin^{3}{u}

    but I don't know how to integrate the second form any more than the first. Can anyone explain how to do this?
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by sinewave85 View Post
    I am stuck!

    \int x^{3}\cos{\tan^{-1}{x}}

    I tried

    \mbox{Let } u=\tan^{-1}{x}, du=\frac{1}{x^{2} +1}

    Then with some substituting,

    \int x^{3}\cos{\tan^{-1}{x}} = \int\sin^{3}{u}

    but I don't know how to integrate the second form any more than the first. Can anyone explain how to do this?

    \int \sin^3(u)du=\int (1-\cos^2(u))\sin(u)du

    Now let w=\sin(u) \implies dw=-\sin(u)du and you are off to the races
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    \int \sin^3(u)du=\int (1-\cos^2(u))\sin(u)du

    Now let w=\sin(u) \implies dw=-\sin(u)du and you are off to the races
    \mbox{But isn't }dw=d(sin(u))=cos(u)du?
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  4. #4
    Member sinewave85's Avatar
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    I scared my helper away

    I still am a little confused.

    If you let w=1-\cos^{2}{u}\implies dw=2\cos{u}\sin{u}\implies u=\cos^{-1}{\sqrt{1-w}}

    Which makes it

    \int\frac{w}{2\sqrt{1-w}}dw

    Then

    z=1-w\implies dz=-1dw \implies w=1-2

    \frac{-1}{2}\int\frac{1-z}{z^{1/2}}dz

    \frac{-1}{2}\int z^{-1/2} - z^{1/2}dz

    And then integrate and back-substitute?
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  5. #5
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    Hi sinewave85

    I can't get : \int x^{3}\cos{\tan^{-1}{x}} ~dx= \int\sin^{3}{u}~du

    Can you tell me how you got it?
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  6. #6
    Member sinewave85's Avatar
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    Quote Originally Posted by songoku View Post
    Hi sinewave85

    I can't get : \int x^{3}\cos{\tan^{-1}{x}} ~dx= \int\sin^{3}{u}~du

    Can you tell me how you got it?
    Sure! (I hope this is legit)

    <br />
\mbox{Let } u=\tan^{-1}{x}\implies du=\frac{1}{x^{2} +1}, u=\tan^{-1}{x}\implies \tan{u}=x

    thus

    <br />
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{x^{3}\cos{u}}{x^{2}+1}du<br />

    <br />
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{tan^{3}{u}\cos{u}}{tan^{2}{u}+1}du<br />

    <br />
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{tan^{3}{u}\cos{u}}{sec^{2}{u}}du<br />

    <br />
\int x^{3}\cos{\tan^{-1}{x}} ~dx= \int\sin^{3}{u}~du<br />

    There you go! I have to admit, this was a real challenge to work through!
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  7. #7
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    Hi sinewave85
    Quote Originally Posted by sinewave85 View Post
    <br />
\mbox{Let } u=\tan^{-1}{x}\implies du=\frac{1}{x^{2} +1}, u=\tan^{-1}{x}\implies \tan{u}=x

    thus

    <br />
\int x^{3}\cos{\tan^{-1}{x}}~dx = \int\frac{x^{3}\cos{u}}{x^{2}+1}du<br />
    You made mistake when substituting dx. You got du=\frac{1}{x^{2} +1}~dx , so dx=(x^2+1)~du
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