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Math Help - New to Integration (and new to forum)

  1. #16
    Math Engineering Student
    Krizalid's Avatar
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    ahhh well, then your substitution will work and 2t\,dt=-2x\,dx\implies t\,dt=-x\,dx.
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  2. #17
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    0_0.... it's so simple. -sqrt(9-x^2) +C

    thanks so much for your help. do you mind if i ask another question?

    i have to have the int of sec (x/2)
    am I on the right track?

    = 1/cos(x/2)
    t=cos(x/2)
    t dt= -1/2sin(x/2)
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  3. #18
    Math Engineering Student
    Krizalid's Avatar
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    no, that substitution won't work.

    that integral is tricky, first put t=\frac x2 so that you only care about integrating \sec t.

    we have \int{\sec t\,dt}=\int{\frac{\sec ^{2}t+\sec t\tan t}{\sec t+\tan t}\,dt}, and then put u=\sec t+\tan t.
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  4. #19
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    you might have to walk me through that one. I understand t dt= sec(t)tan(t), where did the rest of that come from?
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