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# Long trig sub problem

## Video transcript

Let's say we have the indefinite integral of the square root of 6x minus x squared minus 5. And obviously this is not some simple integral. I don't have just, you know, this expression and its derivative lying around, so u-substitution won't work. And so you can guess from just the title of this video that we're going to have to do something fancier. And we'll probably have to do some type of trig substitution. But this immediately doesn't look kind of amenable to trig substitution. I like to do trig substitution when I see kind of a 1 minus x squared under a radical sign, or maybe an x squared minus 1 under a radical sign, or maybe a x squared plus 1. These are the type of things that get my brain thinking in terms of trig substitution. but that doesn't quite look like that just yet. I have a radical sign. I have some x squared, but it doesn't look like this form. So let's if we can get it to be in this form. Let me delete these guys right there real fast. So let's see if we can maybe complete the square down here. So let's see. If this is equal to, let me rewrite this. And if this completing the square doesn't look familiar to you, I have a whole bunch of videos on that. Let me rewrite this as equal to minus 5 minus-- I need more space up here-- minus 5 minus x squared. Now there's a plus 6x, but I have a minus out here. So minus 6 x, right? A minus and a minus will becomes plus 6x. And then, I want to make this into a perfect square. So what number when I add it to itself will be minus 6? Well, it's minus 3 and minus 3 squared. So you take half of this number, you get minus 3, and you square it. Then you put a 9 there. Now, I can't just arbitrarily add nines. Or actually, I didn't add a 9 here. What did I do? I subtracted a 9. Because I threw a 9 there, but it's really a minus 9, because of this minus sign out there. So in order to make this neutral to my 9 that I just threw in there, this is a minus 9. I have to add a 9. So let me add a 9. So plus 9, right there. If this doesn't make complete sense, what I just did, and obviously you have the dx right there, multiply this out. You get minus x squared plus 6x, which are these two terms right there, minus 9, and then you'll have this plus 9, and these two will cancel out, and you'll just get exactly back to what we had before. Because I want you to realize, I didn't change the equation. This is a minus 9 because of this. So I added a 9, so I really added 0 to it. But what this does, it gets it into a form that I like. Obviously this, right here, just becomes a 4, and then this term right here becomes, what? That is x minus 3 squared. x minus 3 squared. So my indefinite integral now becomes the integral, I'm just doing a little bit of algebra, the integral of the square root of 4 minus x minus 3 squared dx. Now this is starting to look like a form that I like, but I like to have a 1 here. So let's factor a 4 out. So this is equal to, I'll switch colors, that's equal to the integral of the radical, and we'll have the 4, times 1 minus x minus 3 squared over 4. I just took a 4 out of both of these terms. If I multiply this out, I'll just get back to that, right there, dx. And now this is starting to look like a form that I like. Let me simplify it even more. So this is equal to the integral, if I take the 4 out, it becomes 2 times the square root of 1 minus, and I can rewrite this as x minus 3, let me write this way. 1 minus x minus 3 over 2 squared dx. And where did I get that 2 from? Well, if I just square both of these, I get x minus 3 squared over 2 squared. Which is x minus 3 over 4. So I have done no calculus so far. I just algebraically rewrote this indefinite integral as this indefinite integral. They are equivalent. But this, all of a sudden, looks like a form that I recognize. I showed you in the last video that cosine squared of theta is just equal to 1 minus sine squared of theta. You could actually do it the other way. You could do sine squared is equal to 1 minus cosine squared. No difference. But they both will work out. But this looks an awful lot like this. In fact, it would look exactly like this if I say that that is equal to sine squared of theta. So let me make that substitution. Let me write that x minus 3 over 2 squared is equal to sine squared of theta. Now, if we take the square root of both sides of that equation, I get x minus 3 over 2 is equal to sine of theta. Now we're eventually, you know where this is going to go. We're eventually going to have to substitute back for theta. So let's solve for theta in terms of x. So theta in terms of x, we could just say, just take the arc sine of both sides of this. You get theta is equal to, right, the arc sine of the sine is just theta. Theta is equal to the arc sine of x minus 3 over 2. Fair enough. Now, to actually do the substitution, though, we're going to have figure out what dx is, we're going to have to solve for x in terms of theta. So let me do that. So we get, if we multiply both sides of the equation by 2, we get x minus 3 is equal to 2 times the sine of theta, or that x is equal to 2 sine of theta plus 3. Now, if we take the derivative of both sides with respect to theta, we get dx d theta, is equal to 2 cosine of theta, derivative of this is just 0. Or we can multiply both sides by d theta, and we get dx is equal to 2 cosine of theta d theta. And we're ready to substitute back into our original indefinite integral. So this thing will now be rewritten as the integral of 2 times the radical, if I can get some space, of 1 minus, I'm replacing this with sine squared of theta. And all that times dx. Well, I just said that dx is equal to this, right here. So dx is equal to 2 cosine of theta d theta. What does this simplify to? This action right here, this is just cosine squared of theta. And we're going to take the square root of cosine squared of theta. So this, the square root of cosine squared of data, this whole term right here, right? That becomes the square root of cosine squared of theta, which is just the same thing as cosine of theta. So our integral becomes, so our integral is equal to, 2 times the square root of cosine squared of theta, so that's just 2 times cosine of theta, times 2 times cosine of theta. That's that one, right there. This is this, and all of this radical sine, that's this, right there. 1 minus sine squared was cosine squared, take the radical, you get cosine squared. And then everything times d theta. Now this is obviously equal to 4 times cosine squared of theta d theta. Which, by itself, is still not an easy integral to solve. I don't know, you know, I can't do U-substitution or anything like that, there. So what do we do? Well, we resort to our good old trig identities. Now, I don't know if you have this one memorized, although it tends to be in the inside cover of most calculus books, or inside cover of most trig books. But cosine squared of theta can be rewritten as 1/2 times 1 plus cosine of 2 theta. And I've proven this in multiple videos. So let's just make this substitution. Let me just replace this thing with that thing. So this integral becomes, it equals, 4 times cosine squared of theta, but cosine squared of theta is this. 4 times 1/2 times 1 plus cosine of 2 theta d theta. Now, this looks easier to deal with. So what is it? 4 times 1/2, that's 2. So my integral becomes the integral of 2 times 1, so it's 2, plus 2 times 2 cosine of 2 theta, all of that d theta. Now, this antiderivative is pretty straightforward. What is this, right here? This is the derivative with respect of theta of sine of 2 theta, right? This whole thing. What's the derivative sine of 2 theta? Take the derivative of the inside, that's 2, times the derivative of the outside, cosine of 2 theta. And this, of course, is the derivative of 2 theta. So this is equal to, the antiderivative of 2 with respect to theta, is just 2 theta, plus the antiderivative of this, which is just sine of 2 theta, and then we have a plus c. And of course, we can't forget that we defined theta, our original antiderivative was in terms of x. So we can't just leave it in terms of theta. We're going to have to do a back substitution. So let's just remember, theta was equal to arc sine of x minus 3 over 2. Let me write that on the side here. Now, if I immediately substitute this theta straight into this, I'm going to get a sine of 2 times arc sine of x minus 3 over 2, which would be correct. And I would have a 2 times arc sine of x minus 3 over 2. That would all be fine, and we would be done. But that's not satisfying. It's not a nice clean answer. So let's see if we can simplify this, so it's only in terms of sine of theta. So when you take the sine of the arc sine, then it just simplifies to x minus 3 over 2. Let me make that clear. So if I can write all of this in terms of sines of theta, because the sine of theta is equal to the sine of the arc sine of x minus 3 over 2. Which is just equal to x minus 3 over 2. So if I can write this in terms of sines of theta, then I can just make this substitution. Sine of theta equals that, and everything simplifies a good bit. So let's see if we can do that. Well, you may or may not know the other identity, and I've proven this as well, that sine of 2 theta, that's the same thing as sine of theta plus theta, which is equal to sine of theta cosine of theta plus, the thetas get swapped around, sine of theta plus cosine of theta, which is equal to, this is just the same thing written twice, 2 sine of theta cosine of theta. Some people have this memorized ahead of time, and if you have to take an exam on trig substitution, it doesn't hurt to have this memorized ahead of time. But let's rewrite this like this. So our indefinite integral in terms of theta, or our antiderivative, became 2 theta, plus, instead of sine of 2 theta, we could write, 2 sine of theta cosine of theta, and of course we have a plus c. Now, I want to write everything in terms of sines of theta, but I have a cosine if theta there. So what can we do? Well, we know we know that cosine squared of theta is equal to 1 minus sine squared of theta, or that cosine of theta is equal to the square root of 1 minus sine squared of theta. Which seems like we're adding complexity to it, but the neat thing is, it's in terms of sine of theta. So let's do that. Let's make the substitution. So our antiderivative, this is the same thing as 2 data plus 2 sine of theta times cosine of theta, which is equal to this, times the square root of 1 minus sine squared of theta, and all of that plus c. Now we're in the home stretch. This problem was probably harder than you thought it was going to be. We know that sine theta is equal to x minus 3 over 2. So let's make that reverse substitution. So we have 2 times theta. This first term is just 2 times theta, right there. So that's 2 times, we can't escape the arc sine. If we just have a theta, we have to say that theta is equal to arc sine of x minus 3 over 2. And then we have plus, let me switch colors, plus 2 sine of theta. Well, that's plus 2 times sine of theta is x minus 3 over 2. So 2 times x minus 3 over 2, and then all of that times the square root of 1 minus sine of theta squared. What's sine of theta? It's x minus 3 over 2 squared. And of course we have a plus c. Let's see if we can simplify this even more, now that we're at the home stretch. So this is equal to 2 arc sine of x minus 3 over 2, plus these two terms, this 2 and this 2 cancel out, plus x minus 3, times the square root of, what happens if we multiply everything here by, let's see, if we take a, so this is equal to 1 minus x minus 3 over 4. x minus 3 squared over 4. This simplification, I should write that in quotes, is taking me longer than I thought. But let's see if we can simplify this even more. If we multiply, let me just focus on this term right here, if we multiply the outside, or say, let's multiply and divide this by 2. So I'll write this as-- so let's just multiply this times 2 over 2. You might say, Sal, why are you doing that? Because I can rewrite this, let me write my whole thing here. So I have 2 arc sine of x minus 3 over 2, and then I have, I could take this denominator 2 right here. So I say, plus x minus 3 over 2. That 2 is that 2 right there. And then I could write this 2 right here as a square root of 4. Times the square root of 4, times the square root of all of this. 1 minus x minus 3 squared over 4. I think you see where I'm going. I'm kind of reversing everything that I did at the beginning of this problem. And maybe I'm getting a little fixated on making this as simple as possible, but I'm so close, so let me finish. So I get 2 times the arc sine of x minus 3 over 2, which I'm tired of writing, plus x minus 3 over 2, and if we bring this 4 in, right, the square root of 4 times the square root of that is equal to the square root of 4 times these things. So it's 4 minus x minus 3 squared, all of that that plus c. And we're at the home stretch. This is equal to 2 times the arc sine of x minus 3 over 2 plus x minus 3 over 2 times the radical of 4 minus, let's expand this, x squared minus 6x plus 9. And then this expression right here simplifies to minus minus, it's 6x minus x squared, and then you have a 4 minus 9 minus 5. Which is our original antiderivative. So finally, we're at the very end. We get the antiderivative is 2 arc sine of x minus 3 over 2 plus x minus 3 over 2 times, time the radical of 6x minus x squared minus 5. That right there is the antiderivative of what this thing that we had at the very top of my little chalkboard, which is right there. So that is equal to the antiderivative of the square root of 6x minus x squared minus 5 dx. And I can imagine that you're probably as tired as I am. My hand actually hurts. But hopefully you find that to be vaguely satisfying. Sometimes I get complaints that I only do easy problems. Well, this was quite a hairy and not-so-easy problem.