How do I integrate..?

Question

-1/2sin2xtan2x

Ok I have now got as far as needing to integrate..

-cos(2x) / 2sin^2(2x)

Answer 1

tan 2x = sin(2x)/cos(2x)

-1/2sin(2x)tan(2x) = - cos(2x)/2 sin(2x)^2

Take t = sin(2x)

dt/dx = 2cos(2x)

int -1/2sin(2x)tan(2x) dx =
= int - cos(2x)/ 2 sin(2x)^2 dx =
= int - cos(2x)/ 2t^2 dx = int -1/4t^2 dt =
= 1/4t +c = 1/4sin(2x) + c

Answer 2

With any integral there's always various methods you can use to get at a solution. With trigonometric functions I often find it usefult to rearrange them into integrals I can solve more easily.
In this case 1/2 sin(2x) tan (2x) it's good to remember tan (2x) is the same as sin(2x)/cos(2x). Putting it in that form you obtain:
1/12sin^2(2x)/cos(2x).
Then knowing sin^2(2x) = 1 - cos^2(2x) we can reformat again to obtain:
1/(2 *cos(2x)) - 1/2 cos(2x).
Here we recognize 1/cos(2x) is also known as sec(2x). Taking the 1/2 out of the integral we have
1/2 INT (sec(2x) - cos(2x)) which is now a very simple integral to solve, giving us a final answer of:
1/4 ln|sec(2x)+tan(2x)| - 1/4 sin (2x) + C

Answer 3

S(-1/2sin2xtan2x)dx
Integration by parts!
Let u=tan2x dv=2sin(2x)
du=2sec^2(2x)dx v=-cos(2x)

uv-Svdu
u*v=tan(2x)*(-cos(2x))=sin2x/cos2x*-cos2x=-sin2x
Svdu=S(-cos(2x)*2sec^2(2x))dx
=S(2sec(2x))dx=ln(sec(x)+tan(x))+c
So, altogether, you have:
S(-1/2sin2xtan2x)dx=-sin(2x)-ln(sec(x)+tan(x))+c