For the first integral, try the substitution .
For the second one, try integration by parts, using the reduction formulas for and .
between 0 and t
Since t is not determined, I'm guessing the second part of the integral is just going to be variables, but I'm having some difficulty getting there.
For likely the same reasons:
between 0 and 2pi (theta)sin(theta)^2 d(theta) is a bugger. halp?
So by the power rule for integration I got:
(u^(n+1)) / (n+1) = u^2/2 and therefore the integration of sin is -cos, plug that into u^2/2 and you get (-cosx^2)/2 (x is easier than theta to show)
except I'm not sure where theta in front of the sin in the original problem and since it is between 0 and 2pi, how to get only the area in that range. Sorry if the questions sound dumb, but in this summer class we do a chapter in a day, so there was 5 min spent on this, and I'm afraid it didn't sink in...
I found it hard to understand what you are trying to say, but I think I got the gist of it.
We'll use x instead of theta. You're right that an antiderivative of the given integral is . To evaluate the area between x = 0 and x = 2pi, then you have to calculate .