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Thread: Integral of (cosx+sinx)^2dx

  1. #1
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    Integral of (cosx+sinx)^2dx

    I did this but not sure if it is correct:
    (cosx+sinx)^2dx=
    (cos^2x+2cosxsinx+sin^2x)=
    ∫(2cosxsinx)=
    2∫cosxsinx=
    2sinx-cosx+C <--------End
    Last edited by Confrassled; Sep 11th 2017 at 12:26 AM.
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  2. #2
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    Re: Integral of (cosx+sinx)^2dx

    $(\cos{x}+\sin{x})^2 = \cos^2{x}+2\sin{x}\cos{x}+\sin^2{x} = 1+2\sin{x}\cos{x} = 1 + \sin(2x)$



    $\displaystyle \int 1 + 2\sin{x}\cos{x} \, dx = x + \sin^2{x} + C$

    or ...

    $\displaystyle \int 1 + 2\sin{x}\cos{x} \, dx = x - \cos^2{x} + C$

    or ...

    $\displaystyle \int 1 + \sin(2x) \, dx = x - \dfrac{\cos(2x)}{2} + C$


    Now a question for you ...

    Why are all three antiderivative forms above valid?
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    Re: Integral of (cosx+sinx)^2dx

    I think i got the answer:
    ∫sin(2x)=
    (using sustitution where u=2x) ∫sin(u)du/2=
    1/2∫sin(u)du=
    1/2-cos(u)=
    1/2-cos(2x)+C <---End
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  4. #4
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    Re: Integral of (cosx+sinx)^2dx

    Failed in 2cosxsinx because it is sin2x, thanks man!
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  5. #5
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    Re: Integral of (cosx+sinx)^2dx

    Quote Originally Posted by Confrassled View Post
    I think i got the answer:
    ∫sin(2x)=
    (using sustitution where u=2x) ∫sin(u)du/2=
    1/2∫sin(u)du=
    1/2-cos(u)=
    1/2-cos(2x)+C <---End
    Thread is close then!
    No ...

    $\displaystyle \int \sin(2x) \, dx = -\dfrac{1}{2}\cos(2x) + C$

    not $\dfrac{1}{2}-\cos(2x) + C$
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