00:01
7 .4, problem 32.
00:03
We're dealing with integrals.
00:04
We're going to be solved with trig substitutions.
00:07
First thing i have to do here, i have to work on this radican.
00:10
So 9 minus 4x squared.
00:12
I need to factor the coefficient of the x -squared term.
00:16
So this is the square root of 4 times 9 -4ths minus x squared.
00:26
Square to 4 is 2.
00:27
So this is 2 times the square root of 9 .4ths.
00:32
Minus x squared.
00:35
So now the substitution to be made is x equal three halves sine of theta, dx is equal to three halves cosine theta d theta and then if you look at this other expression two times nine fourths minus x squared that's two times and you've got nine fourths minus nine -fourths sign squared.
01:18
So this is two times nine -fourths, one minus sine squared of theta.
01:28
So this becomes two times three halves times a cosine of theta...