$\displaystyle \int \frac{10x\cos x-10\sin x}{x^{2}}dx$
How would you go about finding this antiderivative?
Thanks,
Keith
So the quotient rule says $\displaystyle {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}$
So do I just integrate both sides of that with respect to x?
Oh, I see now. So
$\displaystyle u=\sin x$
$\displaystyle {u}'=\cos x$
and
$\displaystyle v=x$
$\displaystyle {v}'=1$
So it'll end up being $\displaystyle \int \frac{10x\cos x-10\sin x}{{x}^2}dx=\frac{10\sin x}{x}+C$?
No.
The function which you want to integrate it is just the derivative of a function.
Find this function.
Use the quoteint rule to find it.
when you want to solve an integral of a function, and you know that derivative of this function, then the integral will be easy.
$\displaystyle \int f'(x) dx = \int \frac{d}{dx}\left(f(x)\right)=f(x)+C$
as an example:
$\displaystyle \int x dx = \int \frac{d}{dx} \left(\frac{1}{2}x^2\right)=\frac{1}{2}x^2+C$.
The function which you want to integrate, is just a derivative of another function.
Try to find it.
I hope you understood what I want to say.
But also what I was trying to get at is trying to get an "inverse quotient rule" that can apply to the quotient rule much as integration by parts applies to the product rule.
so
$\displaystyle {(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^2}$
$\displaystyle \frac{u}{v}=\int\frac{du}{v}-\int\frac{udv}{v^2}$
Or would it be better just to try and recognize the quotient rule?
Is there a rule like this? I have not learned it but I have learned substitution and integration by parts.
Yes. It is.
See:
integrate cos(x)/x - Wolfram|Alpha