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Math Help - Integrate sinx - 2cosx

  1. #1
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    Integrate sinx - 2cosx

    I am trying to use substitution on this indefinite integral.

    Here is what I have so far:

    \int sinx - 2cosx dx

    let u = -cosx

    du/dx = sinx

    du = sinx dx - so I am left with:

    \int 2u du

     = [u^2]

    sub back in

    - cos^2 x

    Where am I going wrong?

    Angus
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  2. #2
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    Re: Integrate sinx - 2cosx

    what you want to integrate is

    \int (\sin(x)-2\cos(x))\;dx

    But you are integrating

    \int -2\sin(x)\cos(x)\;dx

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  3. #3
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    Re: Integrate sinx - 2cosx

    Quote Originally Posted by angypangy View Post
    I am trying to use substitution on this indefinite integral.

    Here is what I have so far:

    \int sinx - 2cosx dx
    = \int sin x dx- 2\int cos x dx

    Or did you mean \int sin(x)(-2cos(x))dx as issacnewton suggests?

    let u = -cosx

    du/dx = sinx

    du = sinx dx - so I am left with:

    \int 2u du

     = [u^2]

    sub back in

    - cos^2 x

    Where am I going wrong?

    Angus
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  4. #4
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    Re: Integrate sinx - 2cosx

    Quote Originally Posted by issacnewton View Post
    what you want to integrate is

    \int (\sin(x)-2\cos(x))\;dx

    But you are integrating

    \int -2\sin(x)\cos(x)\;dx

    Thanks Isaac

    I was revising and read up on substitution assuming I would need to substitute. But of course you are correct. Thanks.
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