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Calculus: Substitution to Integrate Trig Functions

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About this Lesson

  • Type: Video Tutorial
  • Length: 12:44
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 137 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Basics of Integration (14 lessons, $23.76)
Calculus: Integration by Substitution Illustrated (4 lessons, $7.92)

In this lesson, we will work on solving antidifferentiation problems involving composite trigonometric functions using the substitution method to solve for the integral. A composite function is a function that results from applying one function first and then another (e.g. f(g(x))). When these involve trig functions, they look like: find the antiderivative (or integral) of (2x+1)*sin(x^2+x)dx. After using substitution, if you end up with the du-expression being off by a factor of a constant, remember that you can take that constant multiple out of the integral (because of the constant multiple rule of integration and antidifferentiation). Additionally, this lesson will cover the integration of composite functions that involve the trigonometric function secant (in addition to other, more basic, trig functions).

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

Thinkwell
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Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

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Recent Reviews

Nachan_homepage
Great Explanation!
01/20/2009
~ nachan

Professor Burger shows how to find antiderivatives by using substitution. He explains the meaning of sin and how to solve for sin. He explains how to not get confused with long elaborate problems using sin. He also shows how to take a derivative of a problem. Great lesson!

Nachan_homepage
Great Explanation!
01/20/2009
~ nachan

Professor Burger shows how to find antiderivatives by using substitution. He explains the meaning of sin and how to solve for sin. He explains how to not get confused with long elaborate problems using sin. He also shows how to take a derivative of a problem. Great lesson!

The Basics of Integration
Illustrating u-Substitution
Integrating Composite Trigonometric Functions by Substitution Page [1 of 1]
I just ate a KitKat and it was delicious. Oh, of course, that was an endorsement. I just had a chocolate morsel and it was really good, in fact, you may need chocolate morsels to get you through substitution problems. I’ve already seen a whole bunch of them right now, in fact, and mmm, chocolate, even chocolate sort of lingers in your mouth, isn’t it great? Okay, let’s try a few together now. So, again, the theme here is to find the anti-derivatives to these more elaborate functions. Here’s one, the quantity qx + 1, multiplied by sin of (x2 + x)dx. Look at that one. That’s really, really long and complicated. It may seem as though there is no hope in doing this. I want to provide a couple things to you in case there’s any confusion at all.
First of all, we have this quantity as being multiplied by the sin, but I don’t want you to think that this sin is being multiplied by this. Remember, sin is a little function, it’s a trigonometric function. So, sin needs to know what you’re taking the sin of. So, this whole thing is the sin, it’s sin of x2 + x, it’s not multiplication. Sin of that. Now, over here, I’m taking all that here and multiplying it by this quantity. Just in case, you get a little confused sometimes, sometimes students get a little confused and make it sort of this sin times sin times that. This is not multiplication; this is sin of this. If I say to you find the sins, you can’t do anything, you’d say, “Well, find the sin of what”? And I said, “Oh, of power of two”. So, this is what we find the sin of, okay.
Well, how do you proceed? Well, it’s a hard integral; there’s no easy way of doing it that I know of. So, I look at this a little closer and notice something interesting. I noticed that there is an inside, sin of stuff. So, I first see this stuff here, the big blop here, and what about the blop itself? What’s the derivative of the blop? Let’s take the derivative of this in our mind. If I take the derivative of this, we see 2x + 1. And what do you notice? 2x + 1 is basically the multiple that’s sticking out right here. So, when you see that, when you see there’s something inside, that if you take the derivative of it, you get something that’s sitting on the outside somewhere? That cries out for doing the substitution. That cries out for doing the substitution. So, let’s actually try that right now, if it’s crying. If a problem’s crying, what you got to do is help it. So, let’s let you substitute all that inside, all that complicatedness, for one variable. Okay, well, now let’s take the derivative of this and I see that du, and in fact, I’m going to this in one step here, du = (2x + 1)dx. So write down du, dx, and then bring the dx over. I’m going to do that in one step now, if you bring this back, you would see du, dx equals, there’s the derivative.
Okay, well, now we can actually notice that, well, this little bubble right here can be easily replaced by what? Well, that bubble can be replaced by just sin of u. So, let’s right that in. That’s going to be sin of u. So, now something? I’m substituting, that’s why we call this substitution. Sometimes it’s called u substitution or du substitution, so that bubble got replaced by this. Do you see it? Because that’s just here. Now, I can’t integrate stuff with u’s in terms of x’s, so I have to make everything u. But, luckily, from this relationship, I see that this bubble, what’s left, 2x + 1, all multiplied by this dx way out there, see that squiggly bubble. That squiggly bubble is exactly equal to this. So, I can replace everything that’s left by just du. So, I can replace that by du. That’s the squiggly bubble. So, in fact, this integral is equivalent to this one. I now just made a substitution to reduce it. So, I took a hard problem, reduced it to an easier problem in terms of u.
Well, now, to solve this easy problem, that’s not too bad. Because what do you do? Well, what’s the integral of sin? Well, you may be thinking it’s cos, but you got to remember that the derivative of cos is actually minus sin, so I need an extra minus sign in front to make it minus, minus, which is a plus sin, which is here plus a constant. But, of course, that’s not the answer because you brought u into the problem, you have got to take it out. So, I replace all the u’s by stuff with x’s and I see this equals minus cos(x2 + x) + c, and that is the answer to this. The anti-derivative of this function is this. Try it, check it right now, take the derivative of this, you have to use the chain rule, not surprisingly and you’ll get this function back. Notice how I’m finding the right choice of u, taking du, and then seeing how everything else can be recast all in terms of u. That’s the theme here.
Okay, let’s try another one. This one is a little shorter looking. Integral of cos(4x)dx. Well, again I see an inside and an outside. So, how do I proceed? Well, I’m going to let u equal the inside because what do I notice? What’s the derivative of 4x? The derivative of 4x is 4. Now, I don’t see a 4 out here, but I’m allowed to introduce any constant multiple I want. So, it’s okay that we don’t see the 4, because I can just multiply this thing through by 4 and divide by 4 and it won’t change the value. The moral here is that just because the exact derivative doesn’t exist, if you’re off by a constant multiple, then it’s okay. We can actually remedy that, and I’ll show you how. So, let’s take a look at this now and we’ll make a choice here of u. So, let’s let u equal the inside 4x and take the derivative, dudx equals 4 and if I multiply those sides by dx, I would see, du equals 4x. Okay? Is everyone happy? Are you happy? I hope you’re happy. I’m actually pretty happy. I just ate a chocolate morsel.
Okay, so what do I see? I see if this whole bubble, and by the way, I actually do this, I actually circle everything because I have to make sure everything gets replaced by the appropriate thing in terms of u’s, and I’m always afraid I might be a little careless and forget something. So, this whole expression can be replaced by this cos(u), but then I got this bubble here, this squiggly bubble. What do I replace that by? Well, see I can’t just put in du, because du equals 4dx. But see, the constant’s not a problem, I’ll just multiply this equation through by a 4, or divide both sides by the 4. So, in fact I can just modify this, and write here ¼du = dx. Do you see that? I can actually just now bring the 4 over to this side, and now I have just dx by itself. So, if I’m allowed to just multiply by a constant, which I am allowed to do, I can actually modify anything to fit any constants I want. And you’ll see more examples of it, but this is the first example I just divide through. And so, I replace the squiggly bubble, not by du, but ¼ times du, ¼ times du, so I multiply this by ¼ times du. And if you want to simplify this, remember that if you’re integrating something with a constant multiple, you can pull that constant way out in front, so in fact, this just equals ¼ the integral of cos of udu. Remember, a constant you can pull out. Just like when you take a derivative of a constant times a function, the constant can go into a holding, you take the derivative of the function. Same thing with integrals. You have a constant and a function integrated, you can pull the constants out of the holding pattern, take, integrate, and then multiply through with a constant. What’s the integral of cos u? Well, that’s sin u, so this equals ¼ sin u + c, don’t forget that. Well, what’s u? u is 4x let’s plug that in now, ¼ sin of 4x + c, and that’s the answer. That’s the anti-derivative. If you don’t believe me, which I hope you don’t, you take a derivative of this using the chain rule and get this. Let’s do that real real fast. Sin of blop, the derivative of sin of blop is cos of blop, what’s the derivative of 4x, 4, 4/4 is 1. I exactly get that. So, remember, integration’s always great fun, because you can always check your answer, the answer’s given to you.
Okay, let’s try one more example, where you can see this kind of thing happening live. This one is going to look, I think, a little threatening, but don’t panic. This is the integral of x times sec2 of x2. Looks a little scary, I think we got a lot of squares there, sec2 and x2, looks so scary until you remember, the anti-derivative of sec2 is tangent, because the derivative of tangent, we’ve already established in the trig section, is sec2. That is, sec2 here of all this complicated stuff, I wish I could sort of get rid of all that stuff, but notice, that you take the derivative of this inside, we get 2x and that’s a constant basically times this x right here. So, in fact, we can adopt this idea we’ve been developing, so let’s do that right now. Let’s let u = that inside stuff, x2, I take the derivative right now, I see that dudx equals 2x, and if I bring the dx on the bottom over here on this side, I see that over here is a dx, and then how can I re-write this integral? Well, this integral can now be rewritten as how? Well, again what I like to do is circle all the players so this bubble right here gets replaced by sec2u. Now again the hard part in these problems, once you sort of understand the basics, is to figure out what the right choice of u is. Maybe you thought the right choice u might be sec2x2. That would be a good choice, but I don’t think that actually would resolve out into an easier integral. And now, I’m left here with this x times dx, that bubble right there, x times dx. And I don’t have x times dx, however, if I divide both sides by 2, I can make it look that way. So, the thing to do, is to sort of craft this identity, to look like what we want and we can do that as long as we only multiply or divide by constants. And now I have xdx, so I can replace that happily by just ½du. So, this sin squared u was the first bubble and this stuff was that second bubble. So, these two integrals are actually equivalent, and if I rewrite this ½ constant multiple, I can pull out in front, so this equals ½ integral of sec2 udu. What’s the integral of sec2? Well, it’s tangent because the derivative of tangent is sec2 + c, so what’s u? Well, u, I remind you, is ½ tangent and u is just x2 + c. So in fact, this represents the anti-derivative of the original function. Don’t believe me? Take the derivative. Derivative of tangent blop is sec2 blop, peel that up, what’s the derivative of the blop? 2x, but that 2 and that ½ kill themselves and this left with an x alone. So, you can see it.
Okay, so there’s some more examples. These examples have sort of a trigonometric flavor to them, so you see how that works, and up next, we’ll take a look at some other interesting exotic functions and how we can use uv substitutions for those. So, stay with us, have some fun and hopefully get a sense of this general theme. I’ll meet you over there. Bye.

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