00:02
Okay, so this problem can be done very, very quickly, whether you do a or b, because we're doing the integral from 0 to 2 pi of cosine of z over the square root of 4 plus 3 sine of z, dz.
00:30
Okay because if you in part b just you know it just goes from negative pi to pie now explain that here because if you let you equal 4 plus 3 sign of z which would be the method that you would do the substitution you'd have du in this problem would be 3 cosine of z d z so then you know you could you could figure this out that this is one -third to move this over.
01:02
This piece is equal to one -third to you.
01:05
This piece is equal to that.
01:08
But if you notice, if you recognize zero and two -pie are the same values here and here.
01:16
So as you plug in, in matter of fact, they're both zero.
01:20
Sign of zero is zero.
01:22
Sign of two -pie is zero.
01:23
So you'll end up being doing the integral from four to four, of i'll just show you real quick...