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Math Help - integration problem

  1. #1
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    integration problem

    How will you integrate dx
    1+sinxcosx
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  2. #2
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    Quote Originally Posted by gundu View Post
    How will you integrate dx
    1+sinxcosx
    Using the fact that: sin(2 \theta) = 2sin(\theta)cos(\theta), we get:

    \int 1+sin(x)cos(x) dx = \int 1 + \frac{1}{2}sin(2x) dx

    Could you finish it off from here?
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  3. #3
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    Quote Originally Posted by gundu View Post
    How will you integrate dx
    1+sinxcosx
    Remember that \sin{2x} = 2\sin{x}\cos{x}.

    Therefore \sin{x}\cos{x} = \frac{1}{2}\sin{2x}.


    So \int{1 + \sin{x}\cos{x}\,dx} = \int{1 + \frac{1}{2}\sin{2x}\,dx}

     = x - \frac{1}{4}\cos{2x} + C.



    Alternatively you could use a u substitution.

    \int{1 + \sin{x}\cos{x}\,dx} = \int{1\,dx} + \int{\sin{x}\cos{x}\,dx}

     = x + \int{\sin{x}\cos{x}\,dx}.


    Let u = \sin{x} so that \frac{du}{dx} = \cos{x}.


    Then you get

     = x + \int{u\,\frac{du}{dx}\,du}

     = x + \int{u\,du}

     = x + \frac{1}{2}u^2 + C

     = x + \frac{1}{2}\sin^2{x} + C


    It's not difficult to show that the two answers are the same...
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  4. #4
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    thanks

    However the question is actually

    How will you integrate
    1/(1+sinx cosx)

    Sorry for the confusion.
    Last edited by mr fantastic; September 7th 2009 at 11:04 PM. Reason: Merged posts, post moved from another thread, clarification of original post.
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  5. #5
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    Quote Originally Posted by gundu View Post
    thanks

    However the question is actually

    How will you integrate
    1/(1+sinx cosx)

    Sorry for the confusion.
    Learning something from the previous replies, you should be able to re-write the integral as \int \frac{2}{2 + \sin (2x)} \, dx.

    Now read this: Wolfram|Alpha

    Note that the standard Weierstrass substitution is used: PlanetMath: Weierstrass substitution formulas

    You will find several examples if you search these forums using the search tool.

    (There are other ways of doing this integral but they rely on clever algebraic parlor tricks).
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  6. #6
    MHF Contributor red_dog's Avatar
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    \int\frac{1}{1+\sin x\cos x}dx=\int\frac{1}{1+\frac{\sin 2x}{2}}dx=2\int\frac{dx}{2+\sin 2x}

    Now substitute u=\tan x.
    Last edited by mr fantastic; September 8th 2009 at 01:30 AM. Reason: No edit - just flagging the post as having been moved form another thread.
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