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Math Help - integration by part

  1. #1
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    integration by part

    Find \int e^{cos x}sin 2x dx

    what i tried to do was

    \int e^{cos x}sin 2x dx
    =\frac{-1}{sinx}e^{cosx}sin2x-\int \frac{-1}{sinx} e^{cosx}(2)(cos2x) dx
    change sin2x to 2sinxcosx and cos 2x to -2sin^2x+1, then simplify.
    =-2e^{cosx}cosx+4\int (-sinx)(e^{cosx})  dx -2\int \frac{1}{sinx}e^{cosx} dx
    =-2e^{cosx}cosx+4e^{cosx}-2[\frac{-1}{sinx}e^{cosx}(\frac{1}{sinx})-\int(\frac{-1}{sinx}e^{cosx}(-cosec x cotx) dx]

    it's like never going to end.....
    did i make a mistake somewhere?
    or this approach is wrong?
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  2. #2
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    Re: integration by part

    This what I might try. Your integral is

    2\int e^{\cos x} \sin x \cos x dx
    If you let u = \cos x then du = - \sin x\, dx.

    Therefore, your integral transforms to

    -2 \int u \,e^u\,du.

    Then integrate by parts.
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  3. #3
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    Re: integration by part

    Quote Originally Posted by Danny View Post
    This what I might try. Your integral is

    2\int e^{\cos x} \sin x \cos x dx
    If you let u = \cos x then du = - \sin x\, dx.

    Therefore, your integral transforms to

    -2 \int u \,e^u\,du.

    Then integrate by parts.
    i see.
    so, change sin2x to 2sin x cos x first before integrating.
    then, how about the initial approach, will it work eventually?
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