WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to continue our discussion of trigonometric integrals.
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Let us jump right on in.
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Let us recall what the integral of tan(x) looks like.
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Let us work in blue, I really like blue color.
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Let us recall the integral of tan x dx.
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I'm going to rewrite that as the integral of sin x/ cos x dx.
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And then, I'm going to let u equal cos x and du is equal to -sin x dx.
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Sin x dx is equal to –du.
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What we end up actually getting is, this thing is going to be the integral of -du/u.
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Let us come over here = -the integral of du/u = -natlog of u + c is equal – natlog.
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We set that u was cos x = x + c.
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The integral of the tan is –natlog of cos x dx.
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You will also see it written like this, excuse me.
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This can also be written as the natlog sec x + c.
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Either one of these is absolutely fine, I personally prefer that one but where this comes from is this.
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-ln of cos x + c that is the same as 0 - the ln of cos x + c.
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0 can be written as the natlog of 1.
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Natlog of 1 - the natlog of cos x + c is equal to the natlog of this over that 1/ cos x + c,
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which is equal to the natlog of sec x + c.
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That is where it comes from, either one is fine.
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The actual one that you go through, gives you this one.
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This is just another version of it, in case you do not like that negative sign in front, totally a personal choice.
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We have the integral of the sin, the integral of the cos, and now we have done the integral of the tan which we have done before.
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Now let us do the integral of secant.
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Now let us find, and we are doing this just so we actually have a table of the integrals of all of our fundamental trigonometric functions.
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Now let us find the integral of sec x dx.
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This one is just manipulation, that is all it is.
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Again, the manipulation itself is not really important.
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What is important is that we actually find some expression for the integral of the sec(x).
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So that we have an integral of a basic trigonometric function, so that we can use it for other integrals,
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if it happens to come up, which is exactly why we are doing this.
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It equals the integral of sec x × sec x + tan x/ sec x + tan x dx
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is equal to the integral of sec² x + tan x/ sec x + tan x dx.
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We will let u equal to sec x + tan x, du = sec x tan x + sec² x.
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Sec tan + sec dx/u.
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This equals the integral of du/u which = the natlog of u + c which = the natlog of sec x + tan x + c.
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Now we have that one.
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I'm going to leave the integral of cot x dx and the integral of csc x dx to you.
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Do the same way that we just did for these two.
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Now we have the integrals of all six basic trig functions, that is it.
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Now we can continue our discussion of trigonometric integrals.
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Let us try this one, let us go to black, finish the integral, the integral of tan⁵ x.
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Tan⁵ x, I know that tan² x is equal to sec² x – 1.
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How about if I write this as the integral of the tan(x) × tan⁴ (x) dx?
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That is going to equal the integral of the tan(x) × tan² (x)² dx.
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And that is going to equal the integral of tan x × sec² x - 1 dx²,
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that is equal to the integral of tan x × sec⁴ x – 2 sec² x + 1 dx.
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That is going to equal the integral first one and then the second one.
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It is going to equal the integral of tan x sec⁴ x dx - 2 × the integral of tan x sec² x dx + the integral of tan x dx,
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that is why we did the tan and sec in first because when we simplify some other trigonometric integrals,
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one of the integrals that is going to show up is the integral of tan x.
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We wanted to just be able to have something to fill in there as an answer.
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This is our first integral, our second integral, and our third integral.
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Our first integral, the integral of tan x sec⁴ x dx, that is going to equal the integral of tan x sec x.
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I pull out a sec because the sec power is even, I can pull out a sec.
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That leaves me with sec³ x dx.
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I’m going to let u equal the sec(x) and I’m going to let du equal sec x tan x,
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that gives me the integral of u³ du is equal to u⁴/ 4 + c =,
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Let me do this on the next page, I’m actually going to need that.
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That actually equals sec⁴ x/ 4 + c, that is our first integral.
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Our second integral is the integral of tan x sec² x dx.
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We let u equal tan x, du = sec² dx.
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This is actually equal to the integral of u du, that is equal to u²/ 2 + c,
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that is going to be u is tan, it is going to be tan² x/ 2 + c.
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That is our second integral.
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Our third integral, that was the integral of tan x dx.
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We already solved that equals - the natlog of cos x + c.
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We put them all together.
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The integral of tan⁵ x dx is going to end up equaling sec⁴ x/ 4.
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The c’s, I can combine it to 1c, that is not a problem.
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I have this one, and then + tan² x/ 2.
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I have this one, - the natlog of the cos(x) + c.
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There you go, always manipulation, always manipulation.
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It seems like that is all we do in mathematics, is not it?
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That is not all we do, that is a big part of it.
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Now evaluate sec⁵.
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How do we deal with sec⁵?
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In the previous lesson, we dealt with even powers of sec, it also has odd power of sec.
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Let us see what we can do.
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I'm going to rewrite this as sec⁵ x, sec² x dx.
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This is how we deal with an odd power of sec, you try something.
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This is what somebody tried in, it actually ended up working.
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Let us use integration by parts.
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Let me work in blue.
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We will call this u and we will call this dv.
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Let us start all over again.
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The best advice I have ever heard, when you lose your way, go back to the beginning.
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Sec⁵ x, I'm going to write it as the integral of sec³ x sec² x dx.
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There you go, now we are going to use integration by parts.
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Now I go back to blue.
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I’m going to call sec³ x my u, my sec² x is going to be my dv.
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u is equal to sec³ x, dv is equal to sec² x.
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du is equal to 3 sec² x sec x tan x.
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v is just equal to tan x, very nice.
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We know what the integration by parts formula is.
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It is the integral of u dv is equal to uv - the integral of v du.
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Here is my u and here is my v.
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Now it is going to equal, this thing is equal u, u × v.
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It is going to be sec³ x tan(x) - the integral of v du.
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It is going to be the integral of tan x × 3 sec² x sec x tan x dx.
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It is going to be sec³ x tan x - 3 × the integral of sec⁴ x tan² x dx, very nice.
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We have this one already, we just need to deal with this integral right here.
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Let us go ahead and do that.
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I have got 3 × the integral of, we had sec⁴ x tan² x dx, that is equal to 3 × the integral of,
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I have got an even power of sec, I’m going to pull out a sec².
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I’m going to write that as sec² x, sec² x tan² x dx.
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Sec² is equal to 1 + tan², this is actually equal to 3 × the integral of, let me rewrite this.
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The sec², I’m going to write as 1 + tan² x × tan² x × sec² x dx.
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u substitution, I’m going to let u equal tan x.
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I’m going to let du equal sec² x dx.
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Therefore, that gives me 3 × the integral of 1 + u², 1 + tan², u is tan.
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1 + u² × u² × du = 3 × the integral of u² + u⁴ du = 3 × u³/ 3 + u⁵/ 5 + c
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which = 3 ×, u is tan(x), tan³ (x)/ 3 + tan⁵ x/ 5 + c.
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Our final answer, we pull the values that we had.
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The integral of sec⁵ x dx is equal to, we had a sec³ x tan x, that was - this thing.
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-3 tan³ x/ 3 - 3 tan⁵ x/ 5 + c.
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There you go, that is it, just manipulation, that is all we are doing.
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Let us discuss our last set of problems that we might see under the integral sign.
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How to deal with integrals of this type?
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When you have the integral of sin(ax) cos(bx).
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The integral of sin ax sin bx, or the integral of cos ax cos bx.
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You can use the following identities.
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Sin(ax) cos bx, sin and cos where a and b are different.
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Is the same as ½ sin of a - bx + sin a + bx.
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Sin(ax) sin(bx) is ½ cos a - bx - cos a + bx.
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Cos and cos is ½ cos, a - + cos a + b.
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That is it, just using these identities.
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Let us finish off with an example.
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Evaluate the integral cos 5x sin 9x.
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Let us rewrite it to make it look like it was, with sin first, it does not really matter.
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It is sin(9x) × the cos(5x) dx.
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A is equal to 9 and b is equal to 5.
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The identity we are going to be using is sin(ax) cos(bx) = ½ × sin of a - b × x + sin of a + bx.
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This equals the integral, this thing = this thing.
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½ the sin of a – b, 9 – 5, 4x + sin of 9 + 5 is 14x dx = ½ the integral of sin 4x dx + ½ × the integral of sin 14x dx.
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This equals 1/8 - 1/8 the cos of 4x – 1/28.
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½, 1/14, 14 come out.
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Cos(14x) + c, that is it, just using the identities.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.