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, I thought we would actually do some more example problems for trigonometry integrals because they do tend to be tricky.
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I just wanted to throw some examples in there, let us get started.
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The first integral that we are asked to evaluate is the integral sin² cos⁷ dx.
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Remembering or referring back to our summary of things, regarding sin to some power × cos to some power,
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we see that the cos power is odd.
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We are going to actually pull out the cos.
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We are going to write this as the integral of sin² x cos⁶ x cos x dx.
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Now we are going to write this as the integral of sin² x × cos² x³ cos x dx.
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We are going to express cos² x, in terms of sin x.
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Using the Pythagorean identity, sin² x 1 - sin² x³ cos x dx.
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Yes, now I’m going to go ahead and use a u substitution.
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I’m going to let u = sin x.
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I’m going to let du = cos x dx.
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What I end up with is the integral of u² 1 - u²³ du.
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There we go, now let us multiply this out.
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When I multiply this out, I end up with the integral of u² × 1 - 3u² + 3u⁴ – u⁶ du
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=the integral of u² – 3u⁴ + 3u⁶ - u⁸ du.
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=u³/ 3 – 3u⁵/ 5 + 3u⁷/ 7 – u⁹/ 9 + c.
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And then, you plug them back in.
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Now you end up with, final answer is, the integral = sin³ x/ 3 - 3 sin⁵ x/ 5 + 3 sin⁷ x/ 7 – sin⁹ x/ 9 + c.
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Example 2, we want to evaluate x sin² x dx.
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I'm going to rewrite this integral as the integral of x and using that identity with sin²,
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I’m going to write this as ½ × 1 - cos 2x dx.
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This is going to equal the integral of x/ 2 - x cos 2x/ 2 dx = ½ × the integral of x dx - ½ × the integral of x cos x.
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I already took care of the 2 here, x cos x dx.
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This one is going to be ½ × x²/ 2 - this one, I will go ahead and just write -½ the integral of x cos x dx.
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For this one, for this integral, we are going to use integration by parts because we have a function × another function.
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I’m just going to go ahead and use tabular integration.
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Integration by parts, but I’m going to do the quick version of it.
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Never mind, I will do the full integration by parts.
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u is equal to x, dv is equal to cos x, du = 1, the integral cos x = sin x.
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Therefore, this integral ends up being uv.
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The integral of u dv = uv - the integral of v du.
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uv is going to be x sin(x) - the integral of v du - the integral of sin x × 1 dx.
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We have the ½, I put the ½ here and I will put the ½ here.
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We have x²/ 4 - ½ x.
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I think it was sin 2x, was not it?
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Yes, -1/2 × ½ x sin 2x - ½ the integral of sin(2x) dx = x²/ 4 - ¼ x sin(2x).
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I think the strange feeling that I have made a mistake regarding some coefficient to ½ somewhere.
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I think I either had too many or too little.
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But the coefficients are ultimately kind of irrelevant.
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It is the process that we are concerned with.
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The integral of sin, ½ comes out.
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½ × ½ is ¼, another half comes out.
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This is going to be the integral of sin is – cos.
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It is going to be + 1/8 cos(2x) + c.
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There you go, hopefully I have not messed anything up.
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Again, I think there might be an extra ½ somewhere that does not actually belong there, but it is not a problem.
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How does one deal with this one, csc⁴ x/ 5.
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Let us write this as the integral of csc² x/ 5 × csc² x/ 5 dx.
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We know that 1 + cot² x = csc² x.
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We are going to write this as 1 + cot² x/ 5 × csc² x/ 5 dx.
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We will let u equal to cot x/ 5, du = -csc² x/ 5 × 1/5.
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Therefore, -5 du = csc² x/ 5.
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We have the integral of 1 + u² × -5 du.
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It equal -5 × the integral of 1 + u² du which = -5 × u + u³/ 3.
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Our final answer is going to be -5u which is the cot(x)/ 5 - 5 × cot³ (x)/ 5/ 3 + c.
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There you go, that is it, just manipulation.
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1 - tan²/ sec².
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Let us separate this out, I should go back to black.
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This is going to be, this/this and this/this.
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It is the integral of 1/ sec² x dx - the integral of tan²/ sec² x dx.
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This is equal to the integral of cos² x dx - the integral of sin² x/ cos² x/ 1/ cos² x dx.
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Cos² go away, I'm left with the integral of cos² x dx - the integral of sin² x dx.
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This is nice, you can do this as separate using those identities for sin² and cos².
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Or you can recognize that this is just cos² – sin².
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Cos² x – sin² x which is actually equal to the integral of cos 2x.
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Cos² – sin² is double angle identity, cos 2x.
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When you integrate this, integral of cos 2x is going to be ½ sin 2x + c.
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There you go, nice.
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Something that looks really complicated ends up being very simple.
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You will never quite know.
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Something simple, ends up being very complicated.
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Example number 5, evaluate 1/ sin x -1.
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When you see something like this, it is probably best to try to simplify.
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I’m going to multiply by the conjugate of the denominator, just to make it something a little bit more tractable to deal with.
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Let us do 1/ sin x - 1 × sin x + 1 sin x + 1.
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When I multiply all that out, I’m going to end up getting sin x + 1/,
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I will just go ahead and do it, it is not a problem.
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We might as well over here, I’m not going anywhere.
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I’m going to get sin x + 1/ sin² x - 1 which = sin x + 1.
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Sin² x - 1 is equal to -cos² x.
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This is actually equal to the integral of sin x + 1/ -cos² x dx.
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What is the best way to handle this?
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I’m going to separate this out as the integral of sin x/ -cos² x dx + the integral,
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that is fine, let us do it that way.
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1/ -cos² x dx = the integral of sin x/ -cos x.
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I’m going to separate the cos² into cos x.
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Cos x dx, pull this – out, - the integral of, I’m going to write it as sec².
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1/ cos² is sec².
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This is equal to sin x/ cos x.
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I’m going to write is as –tan x.
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-the integral of tan x, 1/ cos x is sec x.
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The integral of sec tan is sec x.
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The integral of sec² is tan x + c.
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There you go, that is it, nice and straightforward.
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The moral of all this is, basically there is no algorithmic approach.
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I know that we summarize some things, when we have even powers of cosines and even powers of sine,
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odd powers of cosine, odd powers of sine, and for the other functions of tan and sec.
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But truth be told, aside from that and aside from some of the identities,
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it really is just a matter of rolling the dice
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and hoping to God that something that you try is actually going to bear some fruit.
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No algorithmic approach, but you are used to that by now.
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This is calculus, this is sophisticated mathematics.
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Not moderately sophisticated, this is sophisticated mathematics and you know it is going to require some experimentation.
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You will go this way, you will go that way.
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You hit a wall, fine, go back to the beginning and start again, not a problem.
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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.