00:01
Okay, we want to evaluate this integral.
00:16
So i'm gonna factor a 16 out of that square root.
00:22
So this becomes 1 quarter integral one to six of dx over x times the square root of x squared minus 3 16ths.
00:41
Okay, so now i'm gonna make a substitution.
00:47
X is gonna be the square root of three divided by four times the secant of let's say u.
01:01
Okay, the reason i want to do this is then x squared minus 3 16ths, which is what's inside the square root is gonna be 3 16ths times one times, excuse me, secant squared of u minus one and secant squared minus one, that's tangent squared.
01:32
Okay, and then dx is secant of u, tangent of u times that square root of three over four du.
02:09
Okay, and then i can, we still have to worry about the limits, but i'll take care of that in a moment.
02:23
Okay, so i'm gonna set this equal to say u of one and u of six, and then our integral is gonna get a root three over four, and then we have a factor out in front of 1 4th, and then from the square.
02:59
And then, so also from the numerator, we get secant of u, tangent of u du.
03:12
And then in the denominator, we have another square root of three over four times the secant of u, that's from the x term.
03:22
And then we get a square root of three over four times the tangent of u, that's from the square root term.
03:31
Okay, so the secants and the tangents all cancel out.
03:37
Okay, so we get, so the 1 4ths cancel, we get root three over three, and then the integral u of one to u of six of du...