Thread: Stuck on trying to find integral of a mixture of trigonometric functions.

Thank so much for all of the help thus far on these problems. This forum has been a life-saver.

So, I have another one ...

So, first things first, here is the problem:
Find the general indefinite integral.

Here are the steps that I have taken so far:

My problem is that I still have the second that I haven’t dealt with. At this point, I am not sure what to do.
Any ideas?
Thank you for your time.
Regards,
Christopher K.

If you multiply the orginal integral through the brackets with the sec outside, you have sec squared + sec*tan, split the sum, so tan + sec + c would be the solution?

Originally Posted by matt.qmar

If you multiply the orginal integral through the brackets with the sec outside, you have sec squared + sec*tan, split the sum, so tan + sec + c would be the solution?

Is that right?

Kind of simple I guess but perhaps this is a compressed form of what everything would look like if I used the chain rule on my solution?

Thanks again.

Regards,
Chris K.