00:01
This problem is from chapter 7, section 2, problem number 12 in the book calculus early transcendental's 8th edition by james stewart.
00:11
Here we have a definite integral from 0 to pi over 2 of parentheses 2 minus sine theta squared.
00:21
So first thing we can do here is just evaluate the square, and we have a 4 minus 4 sine theta plus sine square theta.
00:45
Next, we can use an identity for sine squared, a trigonometric identity, to rewrite this expression here, sine squared, and this becomes 1 minus cosine of 2 theta, all divided by 2.
01:29
So here we could combine this 4 and 1 over 2.
01:33
Add those fractions together first to get 9 over 2, minus 4 sine theta, minus cosine of 2 over 2.
01:58
So we can evaluate each of these three integrals, and it might help for this integral over here, the last integral circled in green, to use a u substitution u equals 2 theta...