Interval, or integral?
I suggest you split it up into
4 sec^2 (theta)
+ 4 sin(theta)/cos^2(theta)
Interval 4sec(theta)(sec(theta)+tan(theta))d(theta)
can someone help me start his problem?
3 answers
You can find the integral of the first one in a table of integrals. The second one is easy if you let u = cos theta and du = -sin theta dtheta
The integral of sec^2 is tan