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Thread: [SOLVED] Trig Integrals

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Trig Integrals

    Hey everyone!

    Integrate.

    \int 59   cos^2 x \cdot sin (2x) dx


    Here is what I have done so far:

    59 \int (\frac{1}{2}(1 + cos 2x) \cdot (2 sinx cosx) dx

    Are those the correct identities to substitute in there?

    Also need some help to go on to the next step of this.

    Thanks y'all!
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  2. #2
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    Hello, mollymcf2009!

    Integrate: . 59\int \cos^2\!x\sin2x\, dx


    Here is what I have done so far: . 59 \int \left(\frac{1 + \cos 2x}{2}\right)(2\sin x\cos x)\,dx
    Don't change both of them.
    There are a few ways to approach this problem.
    Here's one of them . . .

    We have: . 59\int\cos^2\!x(2\sin x\cos x)\,dx \;=\;118\int\cos^3\!x(\sin x\,dx)

    Let: u \,=\, \cos x \quad\Rightarrow\quad du \,=\, -\sin x\,dx \quad\Rightarrow\quad \sin x\,dx \,=\,-du

    Substitute: . 118\int u^3(-du) \;\;=\;\;-118\int u^3\,du \;\;=\;\;-118\,\frac{u^4}{4} + C \;\;=\;\;-\frac{59}{2}u^4 + C

    Back-substitute: . -\frac{59}{2}\cos^4\!x + C

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