# Integration by parts

1. Jun 12, 2009

### nameVoid

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

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

3. The attempt at a solution
u= cos(2x) = > du= -2 sin(2x)
dv=cosx(2x) =>v= 1/2 sin(2x)
?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 12, 2009

### HallsofIvy

Staff Emeritus
And the formula for integration by parts is
$$uv- \int v du$$
which, here, is
$$(1/2)sin(2x)cos(2x)+ \int sin^2(2x)dx$$

Not really an improvement is it? Are you required to use integration by parts? I would use the trig identity $cos^2(u)= (1/2)(1+ cos(2u))$.

3. Jun 12, 2009

### Marksyb

The above identity is probably the easiest way to go, but if you're determined to use integration by parts, try integrating the
$$\int_0^{\frac{\pi}{6}} \sin^2(2x)$$
and see where you end up.