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In this lesson, you will review the integration of sin^3(x) and sin^3(x)*cos^3(x). By substituting for equivalent trigonometric identities, we are able to use the Pythagorean identity for sine and consine (sin^2X+cos^2X = 1) and u-substitution (u-sub) to arrive at the antiderivative of both of these trig expressions. Professor Burger will walk you through the proof of the associated identities derived from this type of manipulation and explain how to recognize other problems which can be solved in the same manner.
This lesson is perfect for review for a CLEP test, mid-term, final, summer school, or personal growth!
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.
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Techniques of Integration
Integrals Involving Powers of Sine and Cosine
Integrals with Powers of Sine and Cosine Page [1 of 2]
Okay, so how can we find the anti-derivative of sin3 x, or cos3 x? So let’s try to integrate that right now. Let’s try to integrate sin3 x dx. Now, actually, since I see an odd power here, there’s a very sneaky thing I can do. See, what I’d like to do is use a udu-substitution. For example, if I had the integral of sin3 x cos x, I’d be in great shape. I could use a udu-substitution. But sadly, there’s no cosine stuff in there. But since this is an odd power, I can do the following little trick. I can actually peel off one of the sines. So I’d have sin2 and then sine. I admit, I’ve done absolutely nothing, this is still sin3. But writing it this way will now empower me to use the Pythagorean theorem formula, which basically tells me that I can now get rid of the sin3 and replace it by a cos2.
Now, let’s just think through the wisdom of this, if any. Why would it be a good thing to convert this sin3 stuff to cos2 stuff? Well, let’s think about it. If this were a cos2 stuff, then I’d see (cos x)2, and notice the inside, which would be cosine, is an object whose derivative is sitting right here, because the derivative of cos = -sin. So that would set me up for a udu-substitution. So the power of using this Pythagorean theorem identity is to actually enable me to convert a sin2 to a cos2, and then I can use the udu-substitution. So let’s see that in action.
So I'm going to do nothing but just insert in place of this – and I want you to look at this and see. What does this equal? It would equal 1 – cos2. So this would be (1 – cos2 x) sin x dx. So I’ve used that identity and what do I do with this? Well, there’s only one game in town and that’s to just take this thing and distribute it so it hits everything. And what happens when I do that? Well, what I see is the integral of sin x – cos2 x sin x dx. Now, this integral right here is pretty easy. What’s the anti-derivative of sin x? Well, it’s –cos x. And why is that? Because take the derivative of –cos x. What you’d see is –(-sin x), because the derivative of cosine is –sin, so that negative and that negative make a positive. Great!
Now, what about this? Well, this actually requires a little udu-substitution, and that was the whole point of making this substitution for sin2 into 1 – cos2, because now if I let u equal, well, the inside – now it might not look like an inside, but, of course, this is really (cos x)2. So let’s let u = cos x, then what would this integral become? Then this integral of cos2 x sin x dx, that would become what? Well, this piece right here would become just u2. But what about all that junk? How could I dismiss that? Well, I dismiss that by taking a derivative. So let’s take the derivative and we see that du = -sin x dx. And that’s not quite what I want. I just want sin x dx, so let me just multiply through by –1, and I’d see –du = sin x dx. And so, in fact, this thing right here, the remaining stuff in the little cloud, that little cloud is just equal to and can be replaced by –du. So that negative sign I could pull out way in front here. There’s the negative sign, don’t forget it, and then I have du. That’s a really easy integral. That’s just going to be . So that’s the integral of this, but the original thing had x’s in it, so let’s now forget the shorthand and replace it by the longhand and put in cos x wherever we see a u. And if we do that, then we’d have the following: we’d see that we’d have a negative and then we’d have, well, that negative is going to be here, and then in place of u, we’d put . And so what we get here is –cos x, and then a negative and a negative is a plus . And so, in fact, you can see that using this technique of first taking off one of the sines and then taking the sin2 and converting it into cosines is a great idea. So whenever you see, in fact, a sine or a cosine function to an odd power, a great technique is to peel one of the factors off and then take the even stuff that’s left and use the Pythagorean theorem identity to convert the even stuff to the opposite even stuff. So a cosine to a sine or a sine to a cosine, and then use a u-substitution.
So, in fact, let’s try another example of this so you can see this in action. Integral sin3 x cos3 x dx. Now that looks really hard, but lets just use the method. If I see either sine or cosine appearing to an odd power, the technique is to rip off one of then, and then you have an even power. In this case, you can do it either way. I could rip off one of the sines, I could rip off one of the cosines, it doesn’t make a difference. Let me, in fact, rip off one of the sines, although you might want to try working this out for yourself again by ripping off one of the cosines. The answer should be the same. So if I rip off one of the sines, what I would see here is sin2 x cos3 x sin x dx. So that’s the still sin3 x and there’s the cos3 x, so nothing’s been done. Now, why is this at all valuable? Well, now I use the Pythagorean theorem identity, which tells me I can now get rid of this sin2 by replacing it by 1 – cos2. And if I do that, here’s what I see. I see the integral of 1 – cos2 x, an then I’ve got this cos3 x, and I’ve got this sin x dx, and this looks scary, because it’s even longer. In fact, look how long that is compared to this. The original question was only a foot long. This is about 17 inches long. It looks like we’re losing ground or, in this case, gaining ground, which is losing ground if you're trying to shrink ground. So are we making progress? Well, the answer is, surprisingly, yes, because if you just follow the procedure now of making the substitution, let’s let u = cos x. If I let u = cos x, then if I differentiate, I see that du = -sin x dx. And if I multiply through by a negative sign, the negative can appear right there.
Now why is this valuable? Well, let’s just go through and convert this now to u’s. If I convert this to u’s, look what I see. This is great! So what does this little bubble right here become? Well, that becomes 1 – u2. This piece becomes u3. And this piece is nothing more than –du. So, in fact, this is nothing more than –du. And now that’s the integral that I have to evaluate. And you may say, “Well, gee, that integral looks certainly smaller, but is it any easier?” And the answer is absolutely, because all I’ve got to do now is just distribute. And if I distribute, I would see minus the integral of u3 – u to what power? Well, I add the exponents and I see a fifth, du. And so what’s that integral? That’s just . That’s it, that’s the whole integral. And now I just have to plug back in what the u stuff is. And what do I see? I see, well, let’s see now. First of all, are you happy with everything? Well, maybe you are. And if that’s so, that’s great. But if you’re not, then you realize that actually I made a classic mistake. Did you see what it was? The classic mistake was that I have a negative sign here and I put the negative sign here, but then what happened here? What I really need to do, as always, is put parentheses around everything, just like that. And so, I’ve got to actually distribute that negative sign here and here. So now, what’s the final answer? Well, the final answer is negative – now, what’s u? Well, u is just cos x, so this is going to be . So, in fact, that is the anti-derivative of this very elaborate-looking integral. So the value of this integral turns out to be this.
And so the moral of the story is if you see sines and cosines being multiplied together and at least one of them has an odd exponent, pick either one you want, if they’re both odd, for example, and rip off one of those pieces. Then you’ll have an even exponent left over. Use the Pythagorean theorem identity and then you’ll have everything in terms of, in this case, cosines with one extra sine left over. So when you make the substitution u = cos, you’ll see that, in this case, -du = sin x dx. And then you convert everything to a really simple integral. So the theme always is take a complicated integral and convert it to an easy integral, evaluate the easy integral and then insert the data.
Okay, so up next, in fact, we’ll take a look at more of these types of questions, where we’re changing the exponents and seeing how to behave. We’ll see what happens if we have at least one odd exponent. We’ll take a look at some other examples up next. I’ll see you there.