1. ## Tough Antiderivative

$\int \frac{10x\cos x-10\sin x}{x^{2}}dx$

How would you go about finding this antiderivative?

Thanks,
Keith

2. Originally Posted by Keithfert488
$\int \frac{10x\cos x-10\sin x}{x^{2}}$

How would you go about finding this antiderivative?

Thanks,
Keith
That is what I called a "tricky" question !
Watch it carefully.

3. So the quotient rule says ${(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}$

So do I just integrate both sides of that with respect to x?

Oh, I see now. So

$u=\sin x$
${u}'=\cos x$

and

$v=x$
${v}'=1$

So it'll end up being $\int \frac{10x\cos x-10\sin x}{{x}^2}dx=\frac{10\sin x}{x}+C$?

4. Originally Posted by Keithfert488
So the quotient rule says ${(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}$

So do I just integrate both sides of that with respect to x?
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.
$\int f'(x) dx = \int \frac{d}{dx}\left(f(x)\right)=f(x)+C$
as an example:
$\int x dx = \int \frac{d}{dx} \left(\frac{1}{2}x^2\right)=\frac{1}{2}x^2+C$.
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.

5. I understand. haha it just came to me. I get it. I edited in the answer.

6. Originally Posted by Keithfert488
So the quotient rule says ${(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^{2}}$

So do I just integrate both sides of that with respect to x?

Oh, I see now. So

$u=\sin x$
${u}'=\cos x$

and

$v=x$
${v}'=1$

So it'll end up being $\int \frac{10x\cos x-10\sin x}{{x}^2}=\frac{10\sin x}{x}$?
Correct.
Do not forget $+C$ and the $dx$.

7. 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

${(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^2}$

$\frac{u}{v}=\int\frac{du}{v}-\int\frac{udv}{v^2}$

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.

8. Originally Posted by Keithfert488
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

${(\frac{u}{v})}'=\frac{v{u}'-u{v}'}{v^2}$

$\frac{u}{v}=\int\frac{du}{v}-\int\frac{udv}{v^2}$

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.
$10\int \frac{cos(x)}{x} dx$ is scary.

9. Originally Posted by General
$10\int \frac{cos(x)}{x} dx$ is scary.
Is that even possible? Isn't it one of those that cannot be represented in terms of elementary functions?

10. Originally Posted by Keithfert488
Is that even possible? Isn't it one of those that cannot be represented in terms of elementary functions?
Yes. It is.
See:
integrate cos(x)/x - Wolfram|Alpha

11. we just integrate by parts:

\begin{aligned}
\int{\frac{x\cos x-\sin x}{x^{2}}\,dx}&=\int{\left( -\frac{1}{x} \right)'(x\cos x-\sin x)\,dx} \\
& =\frac{\sin x-x\cos x}{x}+\cos (x)+k \\
& =\frac{\sin x}{x}+k.
\end{aligned}

12. Originally Posted by Krizalid
we just integrate by parts:

\begin{aligned}
\int{\frac{x\cos x-\sin x}{x^{2}}\,dx}&=\int{\left( -\frac{1}{x} \right)'(x\cos x-\sin x)\,dx} \\
& =\frac{\sin x-x\cos x}{x}+\cos (x)+k \\
& =\frac{\sin x}{x}+k.
\end{aligned}
Wow. That is so much simpler than the way I did it. Thanks! I should have thought of that... =/

13. there're lots of problems involving nasty integrands, but the great key here is to know when to use a good integration by parts.

this is not the most important case, of course.