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Math Help - reverse subsitution (integration)

  1. #1
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    reverse subsitution (integration)

    Hello everyone,
    im having trouble with this question. it says find the integration of this by using reverse subsitution.
    I know we take tan^3x as u and sec^2 and du...but i dont know how to continue..

    reverse subsitution (integration)-maths.jpg

    Thanks.
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  2. #2
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    Quote Originally Posted by Oasis1993 View Post
    Hello everyone,
    im having trouble with this question. it says find the integration of this by using reverse subsitution.
    I know we take tan^3x as u and sec^2 and du...but i dont know how to continue..

    Click image for larger version. 

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    Thanks.
    \int \tan^3(x) \sec^2(x)\,dx

    u = \tan(x) \: \: \rightarrow \: \: du = \sec^2(x)\,dx

    \int \tan^3(x) \sec^2(x)\,dx = \int u^3 \, du

    Which is much easier
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  3. #3
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    thank you!
    so is the answer to this ((tan^3x)^4) / 4?
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  4. #4
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    Almost.

    \frac{1}{4} \tan^{12}(x) + C

    The +C is very important and can cause you to lose the answer mark in an exame


    (\tan^3(x))^4 = ([\tan(x)]^3)^4 = \tan^{12}(x)
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  5. #5
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    thank you!
    Could i also ask you another question regarding reverse subsitution?
    reverse subsitution (integration)-maths.2.jpg

    here again.. 1-x^4 is u and 2x^3 is du?
    but how does this one continue...? :S
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  6. #6
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    Quote Originally Posted by Oasis1993 View Post
    thank you!
    Could i also ask you another question regarding reverse subsitution?
    Click image for larger version. 

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ID:	15834

    here again.. 1-x^4 is u and 2x^3 is du?
    but how does this one continue...? :S
    \int \frac{2x^3}{1-x^4} = 2 \int \frac{x^3}{1-x^4}

    There is a crafty shortcut in this example.

    Set u = 1-x^4. When we differentiate with respect to x we get du = 4x^3 \, dx

    2\int \frac{x^3}{u} \cdot \frac{du}{4x^3}

    This simplifies to \frac{1}{2}\int \frac{du}{u}<br />
which is a standard integral
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