Hell, I don't know the integral of sin^3(phi) off the top of my head, but I can look it up, just like you need to - get a CRC Handbook or some other reference of mathematical functions for a reference if you're going to be doing this for a while.
As long as each variable is independent in this problem, it is mostly a book keeping problem after each step of integration.
Since I'm a nice guy and want to help, the answer from the CRC handbook is:
]sin^3(phi)d(phi) = -(1/3)cos(phi){sin^2(phi) + 2}
This reference
www.efunda.com/math/integrals/show_integrals.cfmgives -
]sin^3(phi)d(phi) = -cos(phi) + cos^3(phi)/3
I'm not going to work through the problem to check that it is the same answer. However, I trust the first answer.
There are several sites on the net that have integration tools, but after looking them over this morning, I would be extremely cautious about their use. Simple integrations all look correct, but due to the complexity of entering the problem statement and any substitutions built into the solver, I get different answers from nearly every solver. Some of them may just be wrong, and I can't figure out how to get some of them to recognize trig functions.
Here's one -
mss.math.vanderbilt.edu/cgi-bin/MSSAgent/~pscrooke/MSS/antiderivative.defan example problem -
www.calc101.com/trig_power_1.htmlIf you enter the engineering profession, there is a good chance that you will need calculus at some point - integrating a complex load distribution over an unuusual shape, for example.
Besides, it's fun!
Remember, it's just adding - at least that is what my Thermo professor told us.
Edited -
Now I need a shower to wash the geekiness off.