Integral of unknown difficulty

[Brandon O]

Integral of unknown difficulty!

Homework Statement

integral of: 1/[4+5sin(2x)] dx

The Attempt at a Solution

U substitution on 2x gives u=2x
then du becomes 2dx so... du=2 dx
after that, i get

1/2 * int of: 1/[4+5sin(u)] du

from here, using v=5sin(u)
then: dv=5cos(u) du

this is where i get stuck. Did i start it off right or did I mess it up here? I don't need the answer to it, just a push in the right direction from the start ><

 

[Cyosis]

This type of integral is typically solved using the substitution $y=\tan(u/2)$.

 

[Brandon O]

My friends and I bashed our heads together long enough to come up with that german guy's name rule.

This is something we haven't done or learned (even though its in chapter 8, which we have done).

Thanks PF ^_^

Our instructor pulled this problem from a... kind of a teacher problem dump website or maybe a book. He didn't check the difficulty on it and put it on the exam anyways. After the initial turn in, he gave them back to the students to do at home... there was a 6 day gap where we could do the problem (as well as the rest of the test).