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Math Help - Trig integrals

  1. #1
    Junior Member symstar's Avatar
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    Trig integrals

    Two problems in this question, the first I was able to do but was getting an incorrect answer and the second I'm just plain stumped on.

    FIRST Q:
    \int\cot^3{x}\csc^3{x}dx

    Here is what I've done so far... I'm coming close but my numbers aren't quite right.

    u=\csc^2{x}
    du=-\cot{x}dx
    \csc{x}=\sqrt{u}

    \int (\csc^2{x}-1)\cot{x}\csc^3{x}dx
    -\int (u-1)u\sqrt{u}du
    -\int (u^{2}-u)u^{1/2}du
    -\int u^{5/2}du + \int u^{3/2}du
    \tfrac{2}{5}u^{5/2}-\tfrac{2}{7}u^{7/2}+C

    ...and subbing in just leaves me with the wrong answer. Where did I go wrong?

    SECOND Q:

    \int\csc{x}dx

    Thought to use \csc{x}=\sqrt{1+\cot^2{x}} or \frac{1}{\sin{x}}, but neither really panned out

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by symstar View Post
    u=\csc^2{x}
    du=-\cot{x}dx
    \csc{x}=\sqrt{u}


    The derivative of \csc^2{x} is not -\cot{x}.

    Pull one factor of \csc{x}\cot{x} and recall that:

    (\csc{x})' = -\csc{x}\cot{x}

    \cot^2{x} + 1 = \csc^2{x}

    As for your second question, multiply it with \frac{\csc{x}+\cot{x}}{\csc{x}+\cot{x}} and see where that gets you.
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  3. #3
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    Quote Originally Posted by symstar View Post
    [snip]
    SECOND Q:

    \int\csc{x}dx

    Thought to use \csc{x}=\sqrt{1+\cot^2{x}} or \frac{1}{\sin{x}}, but neither really panned out

    Thanks in advance.
    The conventional approach is to use the Weierstrass substitution: The Weierstrass Substitution Example
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  4. #4
    Junior Member symstar's Avatar
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    Can't believe I made that mistake... I was thinking of the integral and not derivative. My class isn't up to the Weierstrass substitution yet. Also, I find it funny that a proof for the integral of csc(x) I was viewing stated that the strategy was not obvious... oi!

    Thanks for the help.
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