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  1. #1
    la vache fantôme phantomcow2's Avatar
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    u substitution problem -- integration

    Find the indefinite integral for sec^3x*tanx using u substitution.

    I've done pretty well solving u substitution problems up to this point, but this problem continues to stump me; there is something I am missing.

    let u=tanx
    du = sec^2x*dx

    I figure that maybe I can "break apart" the sec^3x term into sec^2x*secx, giving me the following:
    sec^2x*secx*u*dx

    But du=sec^2dx so substituting gives secx*u*du

    But I can't find the anti derivative of this
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  2. #2
    Look! My Spine! RubenX's Avatar
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    There was a time when I knew this stuff. Sadly I've forgotten it all. My mind is tuned with boolean algebra and logic circuits ATM. Sorry...

  3. #3
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    assuming secx*u*du is correct, the integral to u would be

    0.5*secx*u^2+C

    To see why, just reverse it. Again, not really a math wiz in the formal sense, so please check yourself. Also, this has been ages.

  4. #4
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    Maybe a better way is to write:

    sqrt(1+tan(x)^2)^3*tan(x) dx

    Because then if u=tan(x) you get:

    sqrt(1+u^2)^3*u du --> ((1+u^2)^0.5)^3*u du --> (1+u^2)^1.5*u du

    which eliminates the extra x, which I am not sure what to do about (integrate the result to x aftewards?).

    Finding the anitderivative should be easy from there on, but is has been ages, so you best do that yourself

  5. #5
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    Just a thought:

    (1+u^2)^1.5*u du

    looks like

    h(g(u))'*g(u)'

    with g(u)=1+u^2

    thus h()'=()^1.5 making H()=(()^2.5)/2.5

    ((1+u^2)^2.5)/2.5

    However, the derivative of 1+u^2 is 2u so apparently there was a separate multiplicative term which ended up as 1/u. Then try to compensate for this.

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