Calculus: Integrate Even & Odd Powers of SIN, COS

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Techniques of Integration
Integrals Involving Powers of Sine and Cosine
Integrals with Even and Odd Powers of Sine and Cosine Page [1 of 2]
Okay, so let's take a look at another integral that involves powers of sines and cosines. So here's one. How about the integral of sin^4 x cos^5 x dx? Now that looks like a really complicated integral and we want to find the anti-derivative. Well, I look at this and I notice one thing immediately - one of these exponents is odd. In particular, this one, 5. So I'm going to use this idea about peeling off one of those powers and just having a cosine by itself, and then what remains would actually be an even thing.
Now, let's see why this is going to be valuable. So if I now convert this, I'd have the integral of sin^4 x cos^4 x cos x dx. So I haven't done anything there. That's still cos^5 x. But now I can actually use the Pythagorean identity to change this cosine stuff to sine stuff. Now the step that lets me see that is to notice that, in fact, I could write this as , because that actually is what cos^4 x is. It's just . And now I can use the identity directly and I see that, well, what's cos^2 x? It's 1 - sin^2 x. So if I insert that right in here, what I'd see is the integral . And what's the substitution I should make? I should let u = sin x. By letting u = sin x, I'll see du right here. So once I actually make this change, I let u, in this case, = sin x. Just like in previous examples, if I decompose the sine, if I had an odd power here, then I'd decompose and I'd let u = cosine. In this case, since I'm decomposing the cosine, I let u = sine. And if I let u = sine - in fact, let me just do this up here. So let's let u = sin x, and so du = cos x dx. Now, using those two pieces of information, I can now convert this integral into the following one. In place of sin^4 x, that's just u^4. And instead of 1 - sin^2, I can write for that . And then this stuff here, this sort of pesky-looking stuff, is just du. And that's the power of this simplification, because now all I have to do is integrate that. Now, that actually requires a teeny weeny bit of work, but not really that bad, because all I've got to do is just square this out. So if I just square that out, what do I see? I'm just going to work that out here, so there's u^4. I'll keep that in the picture. And then if I square this out, I see 1 - 2u^2 + u^4. So all I did was just take this and FOIL it out and multiply it by itself. So the first times the first is the 1, the inside terms and the outside terms combine to give me -2u^2, and the last times the last is just u^4. And now, if I distribute the u^4 in, I have u^4 - 2u^6 + u^8 du. And all I've got to do now is integrate that. And so, what's the integral of that? Well, the integral of that is just . So that's pretty easy to do. But what was u? Well, I remind you that u = sine. So wherever I see a u, I can now reduce that shorthand back to the longhand and write in sin x. And if I do that, what I see is the actual integral. And the integral therefore will equal . And so that long answer turns out to be the integral of sin^4 x cos ^5 x. So it's a long answer, but actually not that hard of a process. The thing to do is see if you have an odd exponent in either place. If you seen an odd exponent with cosine, peel off a cosine, then detangle the cosine to sines and let u be sine. If you see an odd exponent on the sine, peel off one of those, make a change of parameters using the Pythagorean theorem to get everything to be in terms of cosines, and then let u = cosine. So that's what we would do if either one of these things is odd.
Question - what if they're both even? Well, how about this? What if I have sin^2 x cos^2 x dx? So now my trick won't work. However, what I'll do now is just immediately apply my squaring formulas. Remember these little double angle formulas. That can reduce the degree. So, for example, if this were actually sin^4, the first thing I would do would be to write this as and use the reduction formula once. And then I would square it out and use the reduction formula again. So you can actually reduce, if you've got even powers in both places, using these formulas. So if we use these formulas, what do I see? Well then this would just equal the integral - in place of sin^2, I'm going to write all that. So that's and in place of the cosine, I'm going to insert dx.
Okay, so I used these identities and now all I've got to do is evaluate that, which may be looks a little bit not pretty, but actually it's not bad at all. First of all, this here and this here, when you multiply them, we get a 4 on the bottom. I can just pull out that in front. It's just a constant multiple, and then I'm left with just that product. And if we multiply that out by FOIL, we see something in this particular example really nice. What we see is just a 1, and then my inside terms give me a -cos 2x and my outside terms give me cos 2x. So they actually add to give 0. They kill each other. And then what am I left with? I'm left with -cox^2 2x dx.
Okay, now how would you integrate this? Well, what I'd do here is I'd first notice that I've got a out in front, and then the integral of 1 is just x. And then I have a - and now I have to integrate cos^2 2x dx. And how do I do that? Well, as we said before, I would just reapply this formula, just reapply the cos^2 thing. So this has a couple of steps to it, but no one step is particularly hard. You just have to carefully go through and continually reduce. The idea is to reduce the degrees of these as you go.
So let's just take a look at that. So if I reduce the degree of that, I would see the integral cos^2 2x dx. That would equal a new integral. And the integral would be 1 + cosine - now cosine of what? So this formula says cos^2 junk = . But in this case, the junk is actually 2x. So I've got to multiply that by 2, which makes it a 4x. So here I have a . And so what does that equal? Well, if you integrate the first part, you're going to see just x. And what about if you integrate the second part? Well, let's see if we can do that. It's going to be a sine - so I have the out in front. And it's going to be a sin 4x. But then when I take the derivative, I get an extra factor of 4, so when I use the udu-substitution, I'd better divide by that 4. And so what I see as the final answer is . That's the integral of that little piece. So that's this little piece right here, so if we combined everything together, what we see is this equals times that integral, which would just compute it. And that's just .
So, in fact, these answers start to get a little bit unwieldy, I think you'll sort of agree. You have to admit that's unwieldy. You can see it because a wield would be round and this is definitely unwieldy, because it's almost rectangular and very unwieldy. So it's very unwieldy. But the power of this method is that originally this seemed like a hopeless integral. And yet, using this reduction technique, we were able to convert it to an integral that was doable. Oh, I admit, several steps to it, so there are many steps, but each step is not that painful. And at the end, we get a very complicated answer, because we were trying to integrate, in fact, a very complicated function.
So the moral is if you see a product of sines and cosines and neither exponent is odd, they're both even, then use these reduction identities, and you might have to use them more than once to reduce the powers. And you can continue this process and keep reducing until you get down to something that can absolutely definitely integrate. All right, I'll see you at the next lecture.