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So I've been sent this definite integral problem and it seemed

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as good as any, and I think the key with this is just to

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see a lot of examples.

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So let's do it.

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This definite integral is from pi over 2 to pi of minus cosine

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squared of x times sin of x dx.

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So before we just chug through the math and do the

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antiderivatives and use the fundamental theorem of calculus

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to evaluate the definite integral, let's think about

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what we're even doing.

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So I've graphed this function right here, minus cosine

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squared of x times sin of x.

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And what we care about, we're defining the definite

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integral between pi over 2, which is roughly here.

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Let me see if I can make this a little bit bigger.

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So between pi over 2 which is right there, and between pi.

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So the definite integral of this function between here and

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here is essentially the area of the curve between the

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curve and the x-axis.

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And since the a curve is below the x-axis here, this area is

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going to be a negative number.

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So that gives us immediately an intuition.

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We should be getting a negative number when we evaluate this

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and just to prove this I actually typed it in up here.

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So let's now evaluate this definite integral.

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Now I'll rearrange some of the terms here just to make it a

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little bit easier to read.

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But the way I always think about it is, well I have

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a cosine and a sin.

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The cosine is squared, so all these crazy things are

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happening to it, so it seems like I could use substitution

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or the reverse chain rules some out here.

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And what was the chain rule?

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The chain rule said if I took the derivative of f of g of x

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that this is equal to f prime of g of x times g prime of x.

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That might completely confuse you, but I just wrote that here

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because we could say, well what if g of x is cosine of x. f

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prime of g of x is the cosine of x squared, and then the

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derivative of g of x or the derivative of cosine

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of x is sin of x.

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Well it's actually minus sin of x, and we have a minus sin

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here, so that works out well too.

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If this confuse you, ignore it.

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Well essentially we're just going to do the same thing,

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but we're going to do it with substitution.

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So let me do it with substitution.

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Let me a erase this if this confuses people.

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I want to do it however it is least confusing to you.

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OK let me erase that.

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Actually, let me do it with substitution, just because the

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way I was just doing is kind of my shortcut back in the

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day when I was a mathlete.

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But it's good to be able to do it with substitution, helps you

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from making careless mistakes.

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So let me rewrite this first of all.

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This is the same thing as the integral from pi over 2 to

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pi of cosine squared of x.

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Actually let me write that as cosine of x squared.

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Same thing, right.

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Times minus sin of x dx.

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And now it should be clearer to you.

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What's the derivative of cosine of x?

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It's minus sin of x.

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So I have a function and it's being squared, and I have its

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derivative, so I can figure out its antiderivative by using

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substitution or the reverse chain rule.

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So let's make a substitution.

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u is equal to cosine of x.

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How did I know to substitute u is equal to cosine of x?

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Well because I say, well the derivative of this

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function is here.

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So when I find du, this whole thing is going to end up

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being du, and let me show that to you.

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So what is du/dx?

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du/dx is equal to minus sin of x.

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That hopefully we've learned already.

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So what is du?

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If we multiply both sides by the differential d of x get du

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is equal to minus sin of x dx.

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So if we look at the original equation, this right here we

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just showed is equal to du, and this right here is what?

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Cosine of x is u.

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That was our original substitution.

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So we have u squared.

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So now let's take the integral.

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And I will arbitrarily switch colors.

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And now this is a very important thing.

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If you're going to do substitution, if we're going to

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say u is equal to cosine of x, we're going to have to actually

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make this substitution on the boundaries.

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Or we could do the substitution and reverse the substitution

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and then evaluate the boundaries, but let's do that.

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So if this is going from x is equal to pi over 2 to pi,

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what is u going from?

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Well when x is equal to pi over 2, u is equal

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to cosine of pi over 2.

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Because u is just cosine of x.

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And then when x is pi, i is going to be cosine of pi.

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And now the fun part.

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Cosine of x squared.

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Well that's just the same thing is u squared.

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And minus sin of x dx, that's the same thing

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is du-- did that here.

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This is pretty straightforward and I'm just going

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to rewrite it.

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What's cosine of pi?

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Cosine of pi is minus 1.

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Cosine of pi over 2, well that's a 0.

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So we have the integral from u is equal to 0 to u is equal to

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negative 1 of u squared du.

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And now this seems like a simple problem.

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So this is equal to the antiderivative of u

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squared, which is fairly straightforward.

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u cubed over 3.

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You could just take the derivative of this,

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you get this.

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All I did is I increased this exponent to get the third, and

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I divided by that exponent.

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Now we're going to have to evaluate it at minus 1 and

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subtract from that, it evaluated at 0.

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So this is equal to minus 1 to the third over 3 minus

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0 over 3, and so this is equal to minus 1/3.

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And we are done.

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And if we look at this area from our original graph, what

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we just solved, as we said the area of the curve between

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here and here is minus 1/3.

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Or if we wanted the absolute area because you can't really

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have a negative area, it's 1/3 but we know it's negative

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because this curve is below the x-axis here.

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And that looks about right, that looks about 1/3.

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I mean if this cube right here is 1, then that

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looks about 1/3.

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The intuition all works out at least.

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So hopefully you found that vaguely useful.

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Actually let me-- since we have a little bit of time.

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Hopefully you understood this, and if you did, don't worry

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about what I'm going to do now, but I want to show you how I

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tend to do it where I just think of it is the

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reverse chain rule.

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It's a little bit quicker sometimes.

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But it's really the same thing as what we just

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did with substitution.

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So if we erase all of this.

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And so if we have this integral right here, what I do is I say,

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well I have cosine of x squared.

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I have cosine of x squared, and then I have minus sin of x.

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This is the derivative of this.

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Since the derivative is here, I can just treat this whole

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thing like an x term.

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So this is the same thing.

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So the antiderivative is cosine of x to the third over 3, and I

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evaluate it at pi and pi over 2.

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And remember, how did I do this?

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What allowed me to treat this cosine of x just like an x or

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like a u when I did it with the substitution?

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Well I had its derivative sitting right here,

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minus sin of x.

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So that's what gave me the license to just take the

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antiderivative, pretend like this cosine of x is just an x,

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is just a u, you could even say, and just take it's

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exponent, raise it by 1 and divide it by 3 and then

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evaluate it from pi to pi over 2.

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So this is equal to cosine of pi cubed over 3 minus cosine

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of pi over 2 cubed over 3.

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This is minus 1 to the third, so this is equal to minus

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1/3 minus this is 0.

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So we get the same answer.

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I just wanted to show you that.

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It's exactly the same with substitution.

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It's just I didn't formally do the substitution, but it's

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the exact same thing.

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Anyway, hope you found that helpful.

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