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Math Help - How do you solve this calculus problem?

  1. #1
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    How do you solve this calculus problem?

    \int x(1-2e^{cotx^{2}})csc^{2}(x^{2})dx

    \int5 e^{\frac{1}{2}lnx}dx

    \int ln e^{\frac{x}{2}}dx


    Can you show how it's done?

    Thank you
    Last edited by honestliar; August 15th 2009 at 08:56 PM.
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  2. #2
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    Hi honestliar

    Do you mean integration ?

    1. let : u=1-2e^{(\cot x^2)}

    2. hint : e^{\ln a}=a

    3. hint : \log a^b = b*\log a
    Last edited by mr fantastic; August 15th 2009 at 11:28 PM. Reason: Merged posts
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  3. #3
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    Quote Originally Posted by honestliar View Post
    \int x(1-2e^{cotx^{2}})csc^{2}(x^{2})dx
    [snip]
    Expand the brackets and tackle 2 separate integrals.

    Just in case a picture helps...



    As usual, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed lines similarly but with respect to the dashed balloon expression. So the triangular network...



    ... is the chain rule.


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    Last edited by mr fantastic; August 18th 2009 at 06:26 AM. Reason: Posts moved from another thread, posts merged, quote added.
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  4. #4
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    Hi honestliar

    1.
    let : u=1-2e^{(\cot x^2)}

    du=-2e^{(\cot x^2)}\; (-\csc^{2}(x^{2})) \; (2x) \; dx= 4x\; e^{(\cot x^2)}\; \csc^{2}(x^{2})\; dx

    Can you continue ?

    P.S: I just realized that Mr. F edited my post. Thx Mr. F
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  5. #5
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    Hi, songoku - I couldn't (yet) make your sub (with parts, maybe?) work for the big one. Hope mine was ok...
    Last edited by mr fantastic; August 18th 2009 at 06:27 AM.
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  6. #6
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    Hi tom@ballooncalculus

    Your method is ok. Hope honestliar understands it

    As for mine, subs. dx into the question and state e^{(\cot x^2)} in term of u.
    Last edited by mr fantastic; August 18th 2009 at 06:28 AM.
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  7. #7
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    D'oh! Cheers for that... and if you don't mind another picture...



    ... where we're doing just as you say, expressing e^{\cot(x^2)} in terms of u (i.e. the dashed balloon expression)...



    ... in order to divide by e^{\cot(x^2)} and thus cancel the extra one appearing in the derivative by-product. I'm using F and G just to point up how the integration with respect to u splits into two.

    Obviously, the lower equals sign depends on several lines of algebra that aren't shown, so the picture's only an overview.


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