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Math Help - Trig Integral

  1. #1
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    Trig Integral

    \displaystyle\int(sin x + cos x)^2dx

    Here are the steps that the book took:
    \displaystyle\int sin^2 x + 2\sin(x) cos(x) + cos^2 x

    \displaystyle\int ((sin^2 x + cos^2 x)2sin(x)cos(x))dx

    \displaystyle\int(1+2sin(x)cos(x))dx

    \displaystyle\int 1dx + \int 2sin(x)cos(x)dx= x+2 \int sin(x)cos(x)dx

    Let u=sin(x); du=cos(x)dx

    \displaystylex+2\int (sin(x)+cos(x))dx=x+2\int u\ du

    \displaystyle = x+2\frac{u^{2}}{2}du+c = x+u^2 + c = x+sin^2(x)+c

    My question is, on this step:

    \displaystyle\int1\ dx + \int(2sinxcosx)dx

    Could I have substituted and come up with this:

    \displaystyle\int1\ dx + \int sin(2x)dx

    \displaystyle = x-\frac{1}{2}cos(2x)+c
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by dbakeg00 View Post
    My question is, on this step:

    \displaystyle\int1\ dx + \int(2sinxcosx)dx

    Could I have substituted and come up with this:

    \displaystyle\int1\ dx + \int sin(2x)dx

    \displaystyle = x-\frac{1}{2}cos(2x)+c
    Yes, that is correct.

    Regards.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Yes, and if you convert it back, you get:

    \begin{array}{cl}x-\frac{1}{2}cos(2x)+c &= x - \dfrac12 (1 -2sin^2(x)) + c \\<br /> <br />
&= x - \dfrac12 + sin^2(x) + 2 \\<br /> <br />
& = x + sin^2(x) - \dfrac12 + c\end{array}

    Where this time, the constant is a different constant from the one the book gave.
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