Calculus: Integrals with Powers of SIN & COS Intro

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Techniques of Integration
Integrals Involving Powers of Sine and Cosine
Introduction to Integrals with Powers of Sine and Cosine Page [1 of 2]
A lot of times in integrals we'll see all sorts of trigonometric functions. And these kind of integrals sometimes look really threatening, because you have all these trig functions and you want to find the anti-derivative of it. And it seems almost hopeless. So I wanted to show you some really neat ways of cracking and opening up these anti-derivatives and actually finding integrals with trig function stuff.
And let's just start off with sort of a not too scary-looking thing. Let's try to find the integral of sin^2 x dx. Now, if it were just sine, that would be pretty easy. That would just be -cosine, because the derivative of -cosine is sine, so the anti-derivative would be -cosine. Sadly, we have the square there. You can imagine another question I could ask and that would be what's the integral of cos^2 x dx? And again, without the square, that's actually a pretty harmless thing to do. The answer would just be sin x + c. But sadly, the square there makes it hard. So how do you deal with an integral that has a sine or a cosine to the power of 2? Well, what I want to do now is try to work out all the angles, in order to really crack the sin^2 x barrier. And actually, there's some really neat formulas that will allow us to crack it, and those formulas are actually useful in a lot of other integrals. So this really is reflective on a more general principle.
So, let's take a look and see if we can actually figure out how to crack sin^2. The first thing I'm going to remind you of is this great identity. It's the Pythagorean identity for sine and cosine. It's that sin^2 + cos^2 = 1. So if you take any x at all and look at sin^2 x and that same x and look at cos^2 and add them up, you always get 1. A fantastic identity.
Now, that identity is one great identity. But there's another really terrific identity that actually revolves around the double angle formulas. So what's the double angle formula stuff again? Well, let me remind you about this really fast. You saw this when you were just a little teeny person back in high school at some point and you thought this was valueless. And, up to this point, maybe you were right. But here we'll actually see a utility of it, So what I want to do is, first of all, remind you of what the cos(2x) equals. Now, cos(2x), it turns out, is equal to cos^2 x - sin^2 x. And notice, by the way, you might say, "Gee, is that 1 or something?" No, the formula is sin^2 x + cos^2 x = 1. But this is the difference, so this is actually something different. And, in fact, this actually is identical - this is actually cos(2x). Now, why is this useful or relevant? Well, if I take this fact and combine it with this fact, I can do one of two things. I could take this relationship and solve it for cos^2, and then replace this cos^2 by stuff with just sin^2. And then I have an identity that would have sin^2 in it. Another thing I could do would be to solve this for sin^2 and take this sin^2 and replace it by what I get here, and then I'll just have cos^2 running around. Either way, I could get run of one of these or the other, using the Pythagorean theorem. So let's actually solve for cosine.
If I solve for cosine, you can see that the Pythagorean identity produces the following: cos^2 x = 1 - sin^2 x. So it's solving for cos^2. I get this. So if I insert this for cos^2, then what do I see? Well, I would see 1 - sin^2 x, and then this - sin^2 x. And putting all that together, what I would see is 1 - 2sin^2 x. So what's the moral of the story? The moral of the story is cos(2x) = 1 - 2sin^2 x. And if I solve this now for sin^2 x, what would I see? Well, sin^2 x would equal - let's just do this now in our heads. So I'm going to bring this whole thing over to this side, so it becomes positive 2sin^2 x. Keep the 1 there and now bring this over to the other side, so I'd see the 1, and then this would be -cos(2x). And then on this side I have that 2 in front. If I divide both sides by 2, I see this fact. And so I've just used the Pythagorean identity and this double angle formula to produce this identity. And that identity is actually really neat, because it allows me to convert sin^2 to just stuff with cosines without any squares in it. So the squares disappear. Now, in fact, if you were to have put in, instead of solving for cosine, if you would have actually solved for sine and plugged in, you would have received a similar identity for cos^2. And that identity actually looks like this: cos^2 x = .
So these are two identities, they look very much alike. The only difference really is the sign. And these two identities allow us to crack the sin^2, because if we return back to the original question, the original question was to figure out this integral. Well, now I can actually use this identity. And if I use that identity, I can replace this dangerous sin^2, which who knows how to deal with, with something that's a little bit more comfortable. So let's actually do that right now and we'll see that sin^2 x = . By the way, a lot of people actually memorize these and that's great. And if you like that, that's fine. But I'll just tell you the truth, just between us that when I think about these formulas, I always have to rederive them, just like I showed you, because I don't know where the sines are and I don't want to memorize things. You know, if you memorize too many things, something really important might skip out of your head. You only have so much space - like maybe your telephone number. And I'd much rather know my telephone number than know those identities, because those identities I can always regenerate just thinking about them. The phone number, there's no hope. If you don't know your phone number, then you can't get it.
So this is great and now what would I do here? Well, I'm going to have to integrate this. I'll break this up into two pieces, just pulling that fraction apart. And so, this integral, that's actually pretty easy. That's just going to be x minus - and then what about that integral? Well, I'll just write that out. There's a out in front, and I see cos(2x). Okay, and what's this integral here? Well, that's going to be a little mini udu-substitution. It's going to be a sin(2x), but unfortunately, if I take the derivative of that, what I see is I'll get an extra factor of 2, which I have to annihilate right now, so I'd better divide by 2. So if I take the derivative, I'd see the derivative sin blop = cos blop, and the derivative of the blop is 2, which will counterbalance with this , and I'm left with just cosine. In fact, that checks. So, in fact, this is the answer + c. And if you want to write this out in a certainly more closed form, you could say x - sin(2x) + c. And now we've actually evaluated this integral that first appeared to be sort of really complicated. And the trick was to crash through with these double angle formulas. In this case, I actually just used this double angle formula. A fun question for you to try is to actually now integrate cos^2 x. Try to integrate cos^2 x by applying this identity and then working through the question that way.
Now, what about a question like this? So now that we're experts, let's see if we can sort of up the ante. How about if I wrote down sin^3 x cos x dx? Now that looks even more threatening than the sin^2, but, in fact, this one is not so bad, because what I see here is that even though it might not look like it, this is just (sin x)^3. So, in fact, the way that I view this is like this. This is just the same thing written differently, but it really emphasizes that there's an inside and then there's an inside. And now, if you notice, the derivative of the sine is actually sitting right here. It's cosine. So, in fact, this is perfectly primed for udu-substitution. Let's let u equal the inside stuff. If I do that, then what does this integral become? This integral becomes u^3. You see how simple that is? That's pretty simple. So that little piece becomes that piece. But what about all this stuff? Well, all this stuff, well, I've got to convert that. So I'll take the derivative . And so if I move the differential over, I see that du = cos x dx. And so, look, this piece right here is exactly equal to just du. And so if I put in du, I get a very simple integral. In fact, the integral of that is . And what was u? Well, u is something that we brought into the problem. The original problem had x's in it, so let's just remove the shorthand and write down the long version, which is . So, in fact, this was a trig question where actually all we had to do was use a u-substitution.
But what about this one? What if I got rid of the cosine and just looked at this? Without the cosine to help us, the udu-substitution no longer is appropriate. So here's a great cliffhanger for you. The question is can we deal with something that's no longer sin^2, but sin^3? I'll see you at the next lecture.