Integration by Parts: Solving for u and v in cos(2x) and cosx(2x)

[nameVoid]

Homework Statement

Homework Statement

The Attempt at a Solution

u= cos(2x) = > du= -2 sin(2x)
dv=cosx(2x) =>v= 1/2 sin(2x)
?

 

[HallsofIvy]

Homework Statement

Homework Statement

The Attempt at a Solution

u= cos(2x) = > du= -2 sin(2x)
dv=cosx(2x) =>v= 1/2 sin(2x)
?

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 $cos^2(u)= (1/2)(1+ cos(2u))$.

 

[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.