Question

Can anyone help with integration?

Answer 1

sure

Answer 2

Thanks! Right now, i'm trying to integrate sin^2(2x). It is the same as (sin(2x))^2 and I 2sin(4x) as a derivative, but I'm not sure where to go from there.

Answer 3

use double angle identity
\[\cos 2 \theta = 1 - 2 \sin^2 \theta\]
\[\rightarrow \sin^2 \theta = \frac{1-\cos 2\theta}{2}\]


now you have
\[ \int\limits \frac{1-\cos 4x}{2} dx\]

u = 4x du = 4dx du/4 = dx
\[\frac{1}{8} \int\limits 1 -\cos u\]
\[= \frac{1}{8}(u -\sin u +C)\]
\[\frac{1}{2} x - \frac{1}{8} \sin 4x + C\]

Answer 4

I'm sorry, I gave you the wrong equation to start. I have to integrate dx/sin^2(2x) which is the same as 1/sin^2(2x)dx.

Answer 5

http://www.wolframalpha.com/input/?i=integrate+csc%5E2+x+dx

Answer 6

Thanks for that!

I figured out how to use the proper identity to make the equation:((1-cos(2x))/2)^-1 but I'm not sure what to use for u and how to substitute.