1. ## 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.

2. ## 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?

3. ## Re: new to integration...

Originally Posted by Nervous
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.

$\displaystyle \int {uv'} = uv - \int {vu'} ~.$

4. ## Re: new to integration...

Originally Posted by Nervous
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*}