00:01
7 .4 .25, the integral 1 over square root of x squared minus 81.
00:07
So in this case, the substitution that we need to make is a secant substitution.
00:13
So i'm going to let x equal 9 times the secant of theta, and then dx is going to be 9, secant theta, tangent theta, d theta, and then x squared minus 81 is going to, be 81 secant squared theta minus 81 and so this just all turns into nine tangent of theta so this integral is one over so you've got nine tangent of theta times d x which is nine secant theta tangent theta d theta you got a nice cancellation right here and so this just becomes the integral of the secant of theta d theta.
01:21
Now taking the antiderivative of the secant, that's going to give you the natural log of the secant of the second of theta plus the tangent of theta plus a constant of integration.
01:34
Now what do we know? what was the original substitution? it was x equal 9 secant of theta.
01:42
So x over 9 is the secant of theta.
01:48
So let's draw this out in a triangle here.
01:52
So here's my angle theta...
