00:01
Let's say i wanted to evaluate the integral i of dx over x squared times root x squared minus 2, using the trigonometric substitution of x equals root 2, secant, theta.
00:18
So what happens if we are to apply this substitution? well, so we're going to want to plug in everywhere we see x.
00:25
We're going to want to plug in root 2, secant theta.
00:28
So this integral is going to become, so we'll do the dx in a second, but it's going to be something over x squared, which is two secant squared theta, two secant squared theta, times the square root, the square root of x squared minus two.
00:53
So x squared is once again this two secant squared theta, two secant squared, theta minus two.
01:08
And this dx, well dx, if we take the derivative of both sides here, dx is going to equal root two times the derivative of secant is secant times tangent.
01:22
So on this side, we're going to have secant theta tan theta d theta.
01:31
And so that's what we're going to replace dx with here.
01:33
So on the top, we have root two, secant, theta, tan theta, d theta.
01:47
So now we want to see what happens here under this radical sign.
01:53
So this root two, we can pull out a root two here.
01:59
And so focusing just on this bottom part, this part, so we have the two secant squared data still.
02:07
And then now we have a times root two times, times the square root of secant squared minus one, which is a trick identity.
02:17
Secant squared minus one is actually just tan squared of theta.
02:24
So we can simplify this root tan squared data to just tan theta, which is great.
02:33
So now our integral will become the integral of root two secant...