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Thread: Integrate (sinx-5cosx)/(sinx+cosx)

  1. #1
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    Integrate (sinx-5cosx)/(sinx+cosx)

    I'm not sure how to integrate this function. I tried converting the numerator into a(sinx+cosx)+b(cosx-sinx), but I couldn't find anything that worked. It didn't really simplify the integrand.
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  2. #2
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    Re: Integrate (sinx-5cosx)/(sinx+cosx)

    Hello, quanta123!

    I found a way, but it's pretty elaborate.
    I'm sure there is a simpler way.

    I used these identities:
    . . . . . \sin^2x \:=\:\frac{1-\cos2x}{2}
    . . . . . \cos^2x \:=\:\frac{1+\cos2x}{2}
    n 2\sin x\cos x \:=\:\sin2x

    \cos^2\!x - \sin^2\!x \:=\:\cos2x


    \text{Integrate: }\:\int\frac{\sin x - 5\cos x}{\sin x - \cos x}\,dx

    Multiply by \frac{\sin x + \cos x}{\sin x + \cos x}:

    . . \frac{\sin x - 5\cos x}{\sin x - \cos x}\cdot\frac{\sin x+\cos x}{\sin x+\cos x} \;=\; \frac{\sin^2\!x + \sin x\cos x - 5\sin x\cos x - 5\cos^2\!x}{\sin^2\!x-\cos^2\!x}

    . . =\;\frac{\sin^2\!x - 4\sin x\cos x - 5\cos^2\!x}{-(\cos^2\!x - \sin^2\!x)} \;=\;\frac{\frac{1-\cos2x}{2} - 2\sin2x - 5\left(\frac{1+\cos2x}{2}\right)}{-\cos2x}

    . . =\;\frac{1-\cos2x - 4\sin2x -5-5\cos2x}{-2\cos2x} \;=\;\frac{-4-4\sin2x - 6\cos2x}{-2\cos2x}

    . . =\;\frac{2 + 2\sin2x + 3\cos2x}{\cos2x} \;=\; \frac{2}{\cos2x} + \frac{2\sin2x}{\cos2x} + \frac{3\cos2x}{\cos2x}

    . . =\;2\sec2x + 2\tan2x + 3


    Got it?
    Thanks from sbhatnagar and cocoboy
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    Re: Integrate (sinx-5cosx)/(sinx+cosx)

    Integrate (sinx-5cosx)/(sinx+cosx)-good-integral.png
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  4. #4
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    Re: Integrate (sinx-5cosx)/(sinx+cosx)

    I think there's a simplest way...

    \displaystyle \int \dfrac{\sin x -5\cos x}{\sin x -\cos x}dx=\int \dfrac{\sin x -cos x -4\cos x}{\sin x - \cos x}dx=\int dx-4\int \dfrac{\cos x}{\sin x -\cos x}dx

    and for the last one integral, using t=\tan \frac{x}{2} will be using simple algebra with a rectangle triangle with t,1,\sqrt{t^2+1} sides
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    Re: Integrate (sinx-5cosx)/(sinx+cosx)

    Hello could you elaborate on how you went from your second step to the third one please. And thank you
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    Re: Integrate (sinx-5cosx)/(sinx+cosx)

    Quote Originally Posted by cocoboy View Post
    Hello could you elaborate on how you went from your second step to the third one please. And thank you
    Why are you resurrecting a 4-year old post? Which second step to which third one are you referring two? Three different people posted solutions.
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    Re: Integrate (sinx-5cosx)/(sinx+cosx)

    Quote Originally Posted by Esteban View Post
    I think there's a simplest way...

    \displaystyle \int \dfrac{\sin x -5\cos x}{\sin x -\cos x}dx=\int \dfrac{\sin x -cos x -4\cos x}{\sin x - \cos x}dx=\int dx-4\int \dfrac{\cos x}{\sin x -\cos x}dx

    and for the last one integral, using t=\tan \frac{x}{2} will be using simple algebra with a rectangle triangle with t,1,\sqrt{t^2+1} sides
    Could you elaborate on how you went from your second step to the third one please.
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