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Evan from Norway has asked me to do another u substitution

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problem, and I like these because it gets my momentum

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going for doing other things that maybe take a little

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bit more preparation.

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And the problem he sent me-- and I hope I pronounced his

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name right-- was the indefinite integral of sin of x over

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the cosine of x squared dx.

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This it also be written as-- he wrote it in email, so I don't

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know how he exactly saw it, but it can also be written as sin

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of x over cosine squared of x.

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Sometimes it's written like that.

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Either way, I like looking at this a bit better.

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It's a little bit less ambiguous.

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But in general you know to do u substitution, or integration by

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substitution when you see something and you see its

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derivative sitting there.

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Right.

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You're like wow, this cosine of x was just an x, or if it was

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just a u, this would be a really easy integral to do.

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We know how to do this integral.

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Let me do it on the side.

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This integral would be easy.

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1 over x squared dx.

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We know how to do that.

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That would just be the antiderivative of x squared.

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This is the same thing as the antiderivative of x to the

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minus 2 dx, and we know how to take the antiderivative

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of something like that.

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You increase the exponent by 1 and then you multiply by

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what your new exponent is.

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So it would be minus x-- I'm sorry.

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You increase your exponent by 1, and then you divide by

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whatever your new exponent is.

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So you would, [? let me do the, ?]

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increasing the exponent x to the minus 2, you'd increase

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the exponent, you'd get x to the minus 1.

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And then when you divide this by minus 1 you get this minus

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out front, and then of course you'd have the plus c.

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If you don't believe it take the derivative.

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Negative 1 times minus 1.

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That's a positive.

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And then you'd decrease the exponent by 1, you

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get x to the minus 2.

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So if we could get it in a form that looks like

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this, we'd be all set.

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And you kind of do see a pattern.

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Where this x is you have a cosine there, and then we have

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cosines derivative there.

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So that's the big clue that we should be using u substitution.

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So let's do that.

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And what we're going to do is we're going to substitute

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u for cosine of x.

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So if we say u is equal to cosine of x, and let's take

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the derivative of u with respect to x.

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So du/dx is equal to what?

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What's the derivative of cosine of x? it's not

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quite sin of x, right?

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It's minus sin of x.

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And then we can multiply both sides by dx, and you get du is

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equal to minus sin of x dx.

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I just multiplied both sides by dx.

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And then up here we have sin of x dx.

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We don't have minus sin of x dx.

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There we have sin of x dx.

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We could have rewritten this top integral, we could have

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rewritten it like this.

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Sin of x dx.

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All of that over cosine of x squared.

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So if we want to substitute for this, here we have a minus.

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Let's multiply both sides of this by a negative 1, and you

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get minus du is equal to sin of x dx.

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And let's see.

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So let's rewrite this original problem.

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Now I know I'm running out of space.

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Let me rewrite it.

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So we know that u is equal to cosine of x, so let's do that.

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So now this integral becomes-- and the denominator, instead of

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cosine of x squared, u is cosine of x.

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That's u, right?

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We made that definition.

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So that's over u squared.

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Cosine of x becomes u.

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And then sin of x dx right up there, what is that equal to?

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Well we just solved for it here.

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That's equal to minus du.

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Sin of x dx is equal to minus du.

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So that we can replace with this, minus du.

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And then of course this has the exact same form as

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this thing right here.

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You could rewrite this, this is equal to let's say

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minus 1 over u squared du.

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I'm just writing it a bunch of different ways.

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Whatever is easier for you to conceptualize.

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The same thing as minus u to the minus 2 du.

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And then here we do the same thing we did up here, although

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now we have a minus out front, that actually makes it

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a little bit cleaner.

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To take the antiderivative, we raise u-- it was to the minus 2

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power, let's raise it to 1 power higher than that-- so

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minus 2 plus 1 is minus 1.

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So it's u to the minus 1 power, and then you want to divide by

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minus 1, and I'll do it explicitly here.

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Minus one.

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And then you had this negative that was sitting out there

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before, so that negative is still going to be there.

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And of course you're going to have a plus c.

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You can view this as a negative 1 or, this negative divided

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by a negative, they're going to cancel out.

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And so you're just left with u to the minus 1 plus c, or 1

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over u plus c is the antiderivative-- oh sorry,

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we're not done yet.

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That's just the antiderivative of this.

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And now we have our substitution to deal with.

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What was our substitution that we started with?

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u is equal to cosine of x.

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So if u is equal to cosine of x, this thing is equal to 1

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over cosine of x plus c is equal to the antiderivative of

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our original problem, which was sin of x over cosine

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of x squared dx.

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There you go.

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See you in the next video.

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