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Example of using trig substitution to solve an indefinite integral. Created by Sal Khan.
Video transcript
Let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared. Of course I have a dx there. So right when I look at that, there's no obvious traditional method of taking this antiderivative. I don't have the derivative of this sitting someplace else in the integral, so I can't do traditional u-substitution. But what I can do is I could say, well, this almost looks like some trig identities that I'm familiar with, so maybe I can substitute with trig functions. So let's see if I can find a trig identity that looks similar to this. Well, our most basic trigonometric identity-- this comes from the unit circle definition-- is that the sine squared of theta plus the cosine squared of theta is equal to 1. And then if we subtract cosine squared of theta from both sides, we get-- or if we subtract sine squared of theta from both sides, we could do either-- we could get cosine squared of theta is equal to 1 minus sine squared of theta. We could do either way. But this, all of a sudden, this thing right here, starts to look a little bit like this. Maybe I can do a little bit of algebraic manipulation to make this look a lot like that. So the first thing, I would like to have a 1 here-- at least, that's how my brain works-- so let's factor out a 3 out of this denominator. So this is the same thing as the integral of 1 over the square root of-- let me factor out a 3 out of this expression. 3 times 1 minus 2/3x squared. I did nothing fancy here. I just factored the 3 out of this expression, that's all I did. But the neat thing now is, this expression looks a lot like that expression. In fact, if I substitute, if I say that this thing right here, this 2/3x squared, if I set it equal to sine squared theta, I will be able to use this identity. So let's do that. Let's set 2/3x squared, let's set that equal to sine squared of theta. So if we take the square root of both sides of this equation, I get the square root of 2 over the square root of 3 times x is equal to the sine of theta. If I want to solve for x, what do I get? And, well, we're going to have to solve for both x and for theta, so let's do it both ways. First, let's solve for theta. If we solve for theta, you get that theta is equal to the arcsine, or the inverse sine, of square root of 2 over square root of 3x. That's if you solve for theta. Now, if you solve for x, you just multiply both sides of this equation times the inverse of this and you get x is equal to-- divide both sides of the equation by this or multiply it by the inverse-- is equal to the square root of 3 over the square root of 2 times the sine of theta. And we were going to substitute this with sine squared of theta, but we can't leave this dx out there. So the derivative of x with respect to theta is equal to square root of 3 over square root of 2. Derivative of this with respect to theta is just cosine of theta, and if we want to write this in terms of dx, we could just write that dx is equal to square root of 3 over the square root of 2 cosine of theta d theta. Now we're ready to substitute. So we can rewrite this expression up here-- I'll do it in this reddish color-- I was using that, let me do it in the blue color. We can rewrite this expression up here now. It's an indefinite integral of-- dx is on the numerator, right? Instead of writing this 1 times dx, I could have just written a dx up here. That could be a dx just like that. You're just multiplying it times dx. So what's dx? dx is this business. I'll do it in yellow. dx is this right here. So it's the square root of 3 over the square root of 2 cosine theta d theta. That's what dx was. Now, the denominator in my equation, I have the square root of 3 times-- now it's 1 minus. Now I said 2/3x squared is equal to sine squared of theta. 93 00:04:47,64 --> 00:04:49,75 Now how can I simplify this? Well, what's 1 minus sine squared of theta? That's cosine squared of theta. So this thing right here is cosine squared of theta. So my indefinite integral becomes the square root of 3 over the square root of 2 cosine theta d theta, all of that over the square root of 3 times the cosine squared of theta. That just became cosine squared of theta. So let's just take the square root of this bottom part. So this is going to be equal to-- I'll do an arbitrary change of colors-- square root of 3 over the square root of 2 cosine of theta d theta, all of that over-- what's the square root of this? It's equal to the square root of 3 times the square root of cosine squared, so times cosine of theta. Now, this simplifies things a good bit. I have a cosine of theta divided by a cosine of theta, those cancel out, so we'll just get 1, and then I have a square root of 3 up here divided by a square root of 3, so those two guys are going to cancel out, so my integral simplifies nicely to 1 over square root of 2 d theta. Or even better, I could write this-- this is just a constant term, I could take it out of my integral-- it equals 1 over the square root of 2 times my integral of just d theta. And this is super easy. This is equal to 1 over the square root of 2 times theta plus c. Plus some constant. I mean, you could say that the integral of this is theta plus c and then you'd multiply the constant times this, but it's still going to be some arbitrary constant. I think you know how to take the antiderivative of this. But are we done? Well, no. We want to know our indefinite integral in terms of x. So now we have to reverse substitute. So what is theta? We figured that out here. theta is equal to arcsine the square root of 2 over the square root of 3x. So our original indefinite integral, which was all of this silliness up here, now that I reverse substitute for theta or put x back in there, it's 1 over the square root of 2 times theta. theta is just this, is just arcsine of square root of 2 over square root of 3 x, and then I have this constant out here, plus c. So this right here is the antiderivative of 1 over the square root of 3 minus 2x squared. So hopefully you found that helpful. I'm going to do a couple of more videos where we go through a bunch of these examples, just so that you get familiar with them.