I'm having a hard time understanding how to do Integrals involving tan^2. I have two problems:

Question

1. Find the integral of (tan^2 y +1)dy
2. Find the integral of (7tan^2 u +15)du

1. My approach to it is to replace the tan^2 y portion of the problem with sec^2 y -1, but it doesn't give me the answer that was given once I've worked it out. Can someone explain to me what I'm doing wrong and how I need to approach these problems?
2. For the latter question, I have no idea how they got the solution 7tanu +8u+C. None as far as tan is concerned and even less of an idea of how they got 8u.

Answer 1

don't forget your trig identities
tan^2+1 = sec^2
∫sec^2 y dy
is easy, right?

7tan^2 u + 15 = 7tan^2 u+7 + 8
= 7sec^2 u + 8
Now we're back to
∫7sec^2 u + 8 du

Answer 2

Thank you very much for your help Steve. I see now where I was going wrong with the problems and feel slightly embarrassed. Thanks again! :)