You could do it using substitution I assume (haven't checked), but here's an approach using some trigonometry tricks.
I'm shifting t to pi/2-t. Of course, the boundaries shift too:
Now we can use and then convert to the half angle using
Now we solved two problems: we introduced a square and we can get rid off the constant 2:
Now we can lose the square root, but watch out that it's the absolute value which comes out of the square root. The problem now is that cos(t/2) becomes negative in our interval for t. But the area under cos(t/2) is the same on the interval [-pi/2,3pi/2] as on [-pi,pi] and on this last one, cos(t/2) is positive so we solved our problem of the absolute value without needing to integrate it.
And now it's easy: