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  1. #1
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    new to integration...

    So im still having trouble integrating functions like this:

    :integration: (x(sin5x))dx

    Am i supposed to use some sort of reverse-product rule? i tried substitution,the only integration tech i know yet.
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    Re: new to integration...

    Hello, Nervous!

    $\displaystyle \int x\sin5x\,dx$

    You say you're "new to integration" . . . How new?

    Are you familiar with Integration by Parts?
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    Re: new to integration...

    Quote Originally Posted by Nervous View Post
    So im still having trouble integrating functions like this :integration: (x(sin5x))dx

    You need integration by parts.

    Look that up in your text material.

    $\displaystyle \int {uv'} = uv - \int {vu'} ~.$
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    Re: new to integration...

    Quote Originally Posted by Nervous View Post
    So im still having trouble integrating functions like this:

    :integration: (x(sin5x))dx

    Am i supposed to use some sort of reverse-product rule? i tried substitution,the only integration tech i know yet.
    Integration by Parts is the most direct method, but if you don't know it, you can always look at the relationship between this function and its derivative.

    $\displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left[ x\sin{(5x)} \right] &= \sin{(5x)} + 5x\cos{(5x)} \end{align*}$

    and

    $\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left[ x\cos{(5x)} \right] &= \cos{(5x)} - 5x\sin{(5x)} \end{align*}$

    So working in reverse, that means we have

    $\displaystyle \displaystyle \begin{align*} \int{\left[ \cos{(5x)} - 5x\sin{(5x)} \right] dx} &= x\cos{(5x)} + c \\ \int{\cos{(5x)}\,dx} - 5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} + c \\ \frac{1}{5}\sin{(5x)} - 5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} + c \\ -5\int{x\sin{(5x)}\,dx} &= x\cos{(5x)} - \frac{1}{5}\sin{(5x)} + c \\ \int{x\sin{(5x)}\,dx} &= \frac{1}{25}\sin{(5x)} - \frac{1}{5}\,x\cos{(5x)} + C \textrm{ where } C = -\frac{1}{5}\, c \end{align*}$
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