WEBVTT mathematics/calculus-ii/murray
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Hello, this is educator.com and today we are going to talk about integration of trigonometric functions.
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The prototypical examples of these integrals is you will have an integral and some power of sine and some power of cosine.
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The important thing to focus on here is what those powers are.
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In particular, which one is odd.
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This is really a game of odds and evens.
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You want one of those powers to be odd.
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If one of them, if they are both even, then the integral gets more difficult and we will talk about that later in the lecture.
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So right now the important thing is to find one of those numbers and to make it the odd or hope that it is odd.
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If they are both odd, then it really does not matter, you can just pick one.
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The one that is odd, what you do with it, is you are going to make a substitution and let U be the other one.
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U is going to be the other one, and dU will be either + or - the one that is odd.
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It seems a little strange so it would be easier to understand if we work through some examples.
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You will see why this strategy works so well.
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Let us take a look at some examples.
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Here is the first example.
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Cos(x) times sin⁴X, dX.
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Of course, cos(x) is the same as cos(x) to the 1.
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So, you look at those numbers 1 and 4, and you see the odd one, that is the odd number.
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What we are going to do is let U be the other one.
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The other one is sin(x) and the point of doing that is dU is then cosin(x) dX.
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So this integral, the cosin, and the dX, become the dU.
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Those two become the dU, sin⁴X just becomes U⁴.
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All of a sudden you have a really easy integral, the integral of U⁴.
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Well, that is just U⁵/5 + C.
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Then you substitute back, and you get sin⁵X + C.
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Just to reiterate there, what made that work was, we looked at which power was odd, and that was the power on cosine.
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Then, we let u be the other one, sin(x)
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The reason we did that is because we had that one power of cosine left over to be the dU.
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So let us try that on a little more complicated example.
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Cos⁴X sinU³X dX.
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Again, we look at those two powers and the odd one is 3.
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We are going to let U be the other one.
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U is cos(x) this time, and so dU, well the derivative of cosin is sin(x) dX
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What we can do now is we can take this sinU³ and we can split that up into a sin²X × sin(x) dX.
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The point here is that we save that sin(x) dX to be the dU.
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Well, we are going to have to attach a negative sign, but that sin(x) dX is going to be the dU.
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Then this sin², we can convert that into 1 - cosin²x.
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That is why we really had to have this odd even pattern going on.
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It is because we save one sin to be the dU, or actually negative dU, and then you can convert sines into cosines two at a time.
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If we save one sine to be dU, that will leave an even number of sines left over.
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We can covert those all into cosines.
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Our integral, after we make all of those substitutions, cos⁴ becomes u⁴.
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The sin² becomes 1 - cos², and then becomes 1 - u⁴.
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Sin(x) dX is dU, or actually negative dU.
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Then we can combine these into one big polynomial.
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Actually, I can bring this negative sign back inside, and make it the integral of u⁴ × u² - 1 dU.
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Combine those into u⁶ - u⁴ dU.
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Then, that is an easy integral because then we can just use the power rule, u⁷/7 - u⁵/5.
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Finally, we can convert that back into u as cos(x).
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So, this is cos⁷X/7 - cos⁵X/5, and of course we have to attach a constant there.
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Again, the key thing here was recognizing those powers.
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We see 4, we see 3, 3 is the odd one, so we want to let u be the other one.
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The other one was cos(x).
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The point of that was it gives you one extra sin(x) left over to be the dU.
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Now, we know what to do if either one of the powers is odd.
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What if you have both powers being even?
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That makes it a trickier problem.
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What we are going to use are the half angle formulas.
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If both sine and cosine are even powers, then we can not use the previous strategy.
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That odd strategy does not work.
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We are going to use the half angle formulas.
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Cos²(x) = 1 + cos(2x)/2
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Sin²X = 1 - cos(2x)/2.
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The point of those is that those are easy to integrate.
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Even if we have to multiply those together a few times, as we will see in the example, it is not too bad to integrate.
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Let us try out an example of that.
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Here we have sin²x cos²x dX.
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We look at those powers, we look for the odd one, and oh no, they are both even.
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So, we are going to use those half angle formulas.
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Remember, sin²x is 1 - cos(2x) over 2.
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Cos²x is 1 + cos(2x)/2.
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Now I am going to pull the two halves out of the integral to get them away.
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So now that is on the outside.
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Now, we have the integral of 1 - cos(2x) × 1 + cos(2x).
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I will multiply those two together and we will get 1 - cos²(2x).
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Well, the 1 is going to be easy to integrate, so I am not going to be worried about that.
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But now we have a cos²(2x).
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How do we integrate that?
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Well, again, that is an even power of cosine.
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We can use the half angle formula again and I am going to write the half angle formula over here.
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But, I am going to use U this time.
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1 + cos(2u)/2 because here we have cos²(2x), but if you think of that as being U, you can convert that into 1/4 the integral of 1 - cos²(u).
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cos²(u) is 1 + cos(2u).
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1 + cos, now 2u if u is 2x, u is 4x.
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This is 1/4 × cos(4x) dX.
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That simplifies a little bit to 1/4 × 1/2 - 1/2 sin(4x) dx.
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I am going to pull those halves out and combine them with the 1/4, and we get 1/8.
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Times the integral of 1 - cos(4x) dX, and that is 1/8.
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Well the integral of 1 is just x.
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The integral of cosine is sine.
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But, because it is 4x, we have to multiply on a 1/4.
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Finally, we get (1/8)x - 1/8 × 1/4 = (1/32)sin(4x) and at the end of these we always add a constant.
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In that one, we had to cope with the fact that there were no odd powers.
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Both were even and we used the half angle formula to resolve that.
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Sometimes after you use the half angle formulas, you find yourself with another even power as we did.
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And so you have to use the half angle formula again to resolve that.
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If you keep on using them, eventually you get down to single powers of cosine and that is something that you can integrate without too much trouble.
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Let us look at another example.
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This one is in terms of secants and tangents.
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That is the other common pattern that you are going to see with trigonometric integrals.
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That is, powers of secant(x), and tangent(x).
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To do this, it is helpful to remember that the derivative of tan(x) is sec²(x).
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The derivative of sec(x) is sec(x)tan(x).
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What we see here is again, looking at these powers, I look at the secant power and I see that that is even.
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The reason that that is significant is that I can separate that out into sec²(x) tan²(x) dx.
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And, I remember that the derivative of tangent is sec²(x).
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If I use the substitution u = tan(x) than dU is sec²(x) dX.
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The point of this is that I can convert this integral into, I have got sec²(x) dX, so that is dU, tan²(x) is u².
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Now what about this other sec², well remember the other pythagorean identity.
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Tan²(x) + 1 is sec²(x).
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Sec²(x) can be written as tan²(x) + 1 which I am going to convert directly into u² + 1.
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This integral converts into u³, sorry, u⁴ + u² dU.
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Again, that is a very easy integral.
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That just integrates to u⁵/5 + u³/3.
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Then we can substitute back, u is tan(x), so that is tan(x⁵/5) + tan(x³/3) + C.
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Let us go back and look at what made that work.
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What made that work again was the even power on sec(x).
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If you have an even power on sec(x), then you can always use this substitution u = tan(x) dU = sec²(x).
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That even power of sec²(x) guarantees you 2 secants to use as the dU,
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and then the other secants you can convert into tangents using the pythagorean identity.
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The other place you can use substitution like this is if the power on tangent is odd, then you can use the other substitution u equals sec(x).
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Then your dU will be sec(x)tan(x) dX.
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The point of that is that you will have an odd tangent.
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Because you have an odd tangent, you will use this pythagorean identity to convert the tangents into secants two at a time.
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Starting with an odd secant you will be left with exactly 1 tangent left and that tangent you can save that to be your dU.
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We will see another example of that later on.