# Homework Help: Integration by parts

1. Jun 26, 2010

### Telemachus

1. The problem statement, all variables and given/known data

$$\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx$$

3. The attempt at a solution
What I did:

$$u=3x$$
$$du=3dx$$
$$dv=cos(\displaystyle\frac{x}{2})$$
$$v=\sin(\displaystyle\frac{x}{2})$$

Then:

$$\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=3x\sin(\displaystyle\frac{x}{2})-\displaystyle\int_{}^{}\sin(\displaystyle\frac{x}{2})3dx=3x\sin(\displaystyle\frac{x}{2})-3\cos(\displaystyle\frac{x}{2})$$

And derive gives me:

$$\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=12\cos(\displaystyle\frac{x}{2})+6x\sin(\displaystyle\frac{x}{2})$$

So I think I'm doing something wrong, but I don't know what. And actually in my first attempt to a solution I got something like: $$\displaystyle\int_{}^{}3x\cos(\displaystyle\frac{x}{2})dx=\displaystyle\frac{1}{3}[x\sin(\displaystyle\frac{x}{2})-\cos(\displaystyle\frac{x}{2})]$$, I arrived to this result by first taking the 3 out of the integral, and passing it to the other side at the end.

Any help will be thanked.

Bye there.

2. Jun 26, 2010

### stevenb

You have the right approach but you made at least 3 silly mistakes. Your expression for "v" is off by a factor of 2 and you are off by a minus sign when you integrate sine. The final step is really silly when you factor out a 3 and get 1/3.

3. Jun 26, 2010

### HallsofIvy

This is incorrect. Let u= x/2. Then du= (1/2)dx so dx= 2du.
$$\int cos(x/2)dx= 2\int cos(u)du= 2 sin(u)+ C= 2sin(x/2)+ C$$
If $dv= cos(x/2)$, then $v= 2sin(x/2)$, not just sin(x/2).

4. Jun 26, 2010

Thanks :)