Originally Posted by

**ecMathGeek** This second integration throws a giant curve ball at you. I've figured out how to do it, it's just that the way to do it is a little tricky.

INT (tan(2x) + cot(2x))^2 dx ... let's turn these into sines and cosines

INT (sin(2x)/cos(2x) + cos(2x)/sin(2x))^2 ... now, get common denominators to add the fractions

INT [((sin(2x))^2 + (cos(2x))^2)/(sin(2x)cos(2x)]^2 dx

INT 1/((sin2x)^2(cos2x)^2) dx

Now here's the catch, we can use (sin2x)^2 + (cos2x)^2 to replace the 1 in the numerator.

You might be thinking, why use that identity in reverse when we already used it to get rid of (sin2x)^2 + (cos2x)^2?

The reason we had to get rid of the (sin2x)^2 + (cos2x)^2 a few steps before is because that whole term was being raised to the second power. If we had multiplied all that out, the problem would get really messy, but there's nothing wrong with replacing 1 with (sin2x)^2 + (cos2x)^2, so that's what we'll do.

INT ((sin2x)^2 + (cos2x)^2)/((sin2x)^2(cos2x)^2) dx

Now let's split up this fraction again. We should get:

INT 1/(cos2x)^2 + 1/(sin2x)^2 dx

INT (sec2x)^2 + (csc2x)^2 ... This we can integrate.