00:02
So we'll start with the integral from 1 to 2, sine 3 theta of rdr, and we'll take the integral of that from 0 to pi over 2, since we're in the first quadrant, d -theta.
00:20
Now, if we start with this interior integral here, we have 1â2r squared, evaluated from 1 to 2 sine 3 theta, and that's going to give us 1â2 times 2.
00:37
2 .5th2 quantity squared minus 1 half times 1 squared.
00:46
Which then gives us 2, sine squared, 3 theta minus 1 half.
00:56
So we'll be taking the integral from 0 to pi over 2 of 2, sine squared, 3 theta, minus 1 1 1â2.
01:12
And so first we'll just break this into two separate integrals.
01:20
So we have the integral of 2, sine squared 3 theta d theta, minus the integral of 1 1â2 d theta.
01:38
Now we can just pull out the constants in each of these.
01:42
So we have 2 times the integral of sine squared 3 theta d theta minus 1 half times the integral of d theta.
01:57
And with this sine squared over here, remember that cosine 2x equals 1 minus 2, sine squared x.
02:20
And so, cosine 2x minus 1 over negative 2 would equal sine squared x, or 1 minus cosine 2x over 2, equals sine squared x...