1. ## Integration question

How can I show that:

$\int \frac{sin(x)}{(2-cos(x))^3}dx = -\frac{1}{2(2-cos(x)^2} + c$

I know that the 2nd half of the equation equates to:

$-\frac{1}{2(4-4cos(x)+cos(x)^2)}$

but how do I integrate the first part of the equation?

Thanks.

2. Use the substitution $u=2-cos(x)$.

3. Originally Posted by Black
Use the substitution $u=2-cos(x)$.
Okay so now we have:

$\int \frac{sin(x)}{(u)^3}dx$

Does this now make:

$\int \frac{sin(x)}{u}$

$\frac{-cos(x)}{(u)^3}$

4. u=2-cos(x)
du=?

subsitute u and du in the equation, then find the integral

5. Originally Posted by Zocken
u=2-cos(x)
du=?

subsitute u and du in the equation, then find the integral
$sin(x)$

Isn't this now $sin(x)^{3}$ because of the power?

6. I didn't say substitute u for dx, I said substitute u *and* dx. Do you know what u substitution is?

7. Originally Posted by Zocken
I didn't say substitute u for dx, I said substitute u *and* dx. Do you know what u substitution is?
To be honest I've only been doing this for a few weeks so I'm not sure that's why I'm after help where I can learn from others.

8. you have this equation

$

\int \frac{sin(x)}{(2-cos(x))^3}dx
$

you have u=2-cos(x) du=?(which you found)

do you know how to substitute (you don't need rocket science for this)?

9. $\int \frac{du}{u}dx$

Is this now equal to:

$ln(u)+c$

10. there's not an ln in the equation you're trying to prove so I think not.

11. $u= 2-cosx \Rightarrow \frac{du}{dx} = sinx \Rightarrow dx = \frac{du}{sinx}$

Now substitute $2-cosx$ with $u$ and $dx$ with $\frac{du}{sinx}$. What's the result?

12. let $u=2-cos(x)$
$du=sin(x)=-cos(x)$

$=\int\frac{du}{(u)^3}dx$

$=\frac{-1}{2(2-cos(x)^2)} + c$

God do you know when you stare too long at something, well that was me!

It's 0045 here & I need my bed!

13. You still didnt substitute dx. And $du = sinx = -cosx$? What does this mean?

Carefully read (a few times if neccessary) the instructions from my last post. You have both dx and u. There should be no dx in the integral after you're done substituting.