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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- Dec 2nd 2012, 01:44 PMNervousnew 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. - Dec 2nd 2012, 01:55 PMSorobanRe: new to integration...
Hello, Nervous!

Quote:

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

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

Are you familiar with*Integration by Parts?*

- Dec 2nd 2012, 01:56 PMPlatoRe: new to integration...
- Dec 2nd 2012, 03:15 PMProve ItRe: new to integration...
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

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