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Let's say we have the indefinite integral of the

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square root of 6x minus x squared minus 5.

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And obviously this is not some simple integral.

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I don't have just, you know, this expression and its

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derivative lying around, so u-substitution won't work.

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And so you can guess from just the title of this video that

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we're going to have to do something fancier.

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And we'll probably have to do some type of trig substitution.

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But this immediately doesn't look kind of amenable

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to trig substitution.

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I like to do trig substitution when I see kind of a 1 minus x

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squared under a radical sign, or maybe an x squared minus

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1 under a radical sign, or maybe a x squared plus 1.

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These are the type of things that get my brain thinking in

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terms of trig substitution.

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but that doesn't quite look like that just yet.

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I have a radical sign.

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I have some x squared, but it doesn't look like this form.

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So let's if we can get it to be in this form.

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Let me delete these guys right there real fast.

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So let's see if we can maybe complete the square down here.

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So let's see.

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If this is equal to, let me rewrite this.

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And if this completing the square doesn't look familiar

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to you, I have a whole bunch of videos on that.

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Let me rewrite this as equal to minus 5 minus-- I need more

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space up here-- minus 5 minus x squared.

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Now there's a plus 6x, but I have a minus out here.

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So minus 6 x, right?

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A minus and a minus will becomes plus 6x.

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And then, I want to make this into a perfect square.

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So what number when I add it to itself will be minus 6?

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Well, it's minus 3 and minus 3 squared.

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So you take half of this number, you get minus

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3, and you square it.

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Then you put a 9 there.

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Now, I can't just arbitrarily add nines.

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Or actually, I didn't add a 9 here.

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What did I do?

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I subtracted a 9.

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Because I threw a 9 there, but it's really a minus 9, because

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of this minus sign out there.

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So in order to make this neutral to my 9 that I

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just threw in there, this is a minus 9.

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I have to add a 9.

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So let me add a 9.

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So plus 9, right there.

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If this doesn't make complete sense, what I just did, and

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obviously you have the dx right there, multiply this out.

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You get minus x squared plus 6x, which are these two terms

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right there, minus 9, and then you'll have this plus 9, and

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these two will cancel out, and you'll just get exactly back

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to what we had before.

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Because I want you to realize, I didn't change the equation.

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This is a minus 9 because of this.

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So I added a 9, so I really added 0 to it.

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But what this does, it gets it into a form that I like.

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Obviously this, right here, just becomes a 4, and then this

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term right here becomes, what?

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That is x minus 3 squared.

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x minus 3 squared.

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So my indefinite integral now becomes the integral, I'm just

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doing a little bit of algebra, the integral of the square root

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of 4 minus x minus 3 squared dx.

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Now this is starting to look like a form that I like, but

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I like to have a 1 here.

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So let's factor a 4 out.

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So this is equal to, I'll switch colors, that's equal to

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the integral of the radical, and we'll have the 4, times 1

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minus x minus 3 squared over 4.

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I just took a 4 out of both of these terms.

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If I multiply this out, I'll just get back to that,

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right there, dx.

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And now this is starting to look like a form that I like.

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Let me simplify it even more.

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So this is equal to the integral, if I take the 4 out,

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it becomes 2 times the square root of 1 minus, and I can

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rewrite this as x minus 3, let me write this way.

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1 minus x minus 3 over 2 squared dx.

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And where did I get that 2 from?

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Well, if I just square both of these, I get x minus 3

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squared over 2 squared.

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Which is x minus 3 over 4.

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So I have done no calculus so far.

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I just algebraically rewrote this indefinite integral as

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this indefinite integral.

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They are equivalent.

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But this, all of a sudden, looks like a form

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that I recognize.

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I showed you in the last video that cosine squared of theta

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is just equal to 1 minus sine squared of theta.

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You could actually do it the other way.

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You could do sine squared is equal to 1 minus

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cosine squared.

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No difference.

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But they both will work out.

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But this looks an awful lot like this.

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In fact, it would look exactly like this if I say that that is

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equal to sine squared of theta.

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So let me make that substitution.

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Let me write that x minus 3 over 2 squared is equal

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to sine squared of theta.

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Now, if we take the square root of both sides of that equation,

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I get x minus 3 over 2 is equal to sine of theta.

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Now we're eventually, you know where this is going to go.

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We're eventually going to have to substitute back for theta.

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So let's solve for theta in terms of x.

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So theta in terms of x, we could just say, just take the

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arc sine of both sides of this.

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You get theta is equal to, right, the arc sine of

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the sine is just theta.

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Theta is equal to the arc sine of x minus 3 over 2.

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Fair enough.

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Now, to actually do the substitution, though, we're

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going to have figure out what dx is, we're going to have to

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solve for x in terms of theta.

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So let me do that.

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So we get, if we multiply both sides of the equation by 2, we

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get x minus 3 is equal to 2 times the sine of theta,

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or that x is equal to 2 sine of theta plus 3.

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Now, if we take the derivative of both sides with respect to

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theta, we get dx d theta, is equal to 2 cosine of theta,

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derivative of this is just 0.

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Or we can multiply both sides by d theta, and we get dx is

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equal to 2 cosine of theta d theta.

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And we're ready to substitute back into our original

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indefinite integral.

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So this thing will now be rewritten as the integral of 2

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times the radical, if I can get some space, of 1 minus,

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I'm replacing this with sine squared of theta.

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And all that times dx.

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Well, I just said that dx is equal to this, right here.

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So dx is equal to 2 cosine of theta d theta.

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What does this simplify to?

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This action right here, this is just cosine squared of theta.

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And we're going to take the square root of

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cosine squared of theta.

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So this, the square root of cosine squared of data, this

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whole term right here, right?

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That becomes the square root of cosine squared of theta, which

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is just the same thing as cosine of theta.

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So our integral becomes, so our integral is equal to, 2 times

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the square root of cosine squared of theta, so that's

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just 2 times cosine of theta, times 2 times cosine of theta.

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That's that one, right there.

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This is this, and all of this radical sine, that's

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this, right there.

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1 minus sine squared was cosine squared, take the radical,

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you get cosine squared.

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And then everything times d theta.

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Now this is obviously equal to 4 times cosine squared

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of theta d theta.

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Which, by itself, is still not an easy integral to solve.

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I don't know, you know, I can't do U-substitution or

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anything like that, there.

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So what do we do?

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Well, we resort to our good old trig identities.

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Now, I don't know if you have this one memorized, although it

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tends to be in the inside cover of most calculus books,

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or inside cover of most trig books.

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But cosine squared of theta can be rewritten as 1/2 times

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1 plus cosine of 2 theta.

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And I've proven this in multiple videos.

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So let's just make this substitution.

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Let me just replace this thing with that thing.

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So this integral becomes, it equals, 4 times cosine

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squared of theta, but cosine squared of theta is this.

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4 times 1/2 times 1 plus cosine of 2 theta d theta.

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Now, this looks easier to deal with.

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So what is it?

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4 times 1/2, that's 2.

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So my integral becomes the integral of 2 times 1, so it's

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2, plus 2 times 2 cosine of 2 theta, all of that d theta.

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Now, this antiderivative is pretty straightforward.

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What is this, right here?

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This is the derivative with respect of theta of

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sine of 2 theta, right?

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This whole thing.

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What's the derivative sine of 2 theta?

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Take the derivative of the inside, that's 2, times the

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derivative of the outside, cosine of 2 theta.

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And this, of course, is the derivative of 2 theta.

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So this is equal to, the antiderivative of 2 with

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respect to theta, is just 2 theta, plus the antiderivative

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of this, which is just sine of 2 theta, and then

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we have a plus c.

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And of course, we can't forget that we defined theta, our

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original antiderivative was in terms of x.

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So we can't just leave it in terms of theta.

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We're going to have to do a back substitution.

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So let's just remember, theta was equal to arc

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sine of x minus 3 over 2.

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Let me write that on the side here.

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Now, if I immediately substitute this theta straight

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into this, I'm going to get a sine of 2 times arc sine of x

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minus 3 over 2, which would be correct.

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And I would have a 2 times arc sine of x minus 3 over 2.

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That would all be fine, and we would be done.

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But that's not satisfying.

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It's not a nice clean answer.

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So let's see if we can simplify this, so it's only in

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terms of sine of theta.

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So when you take the sine of the arc sine, then it just

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simplifies to x minus 3 over 2.

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Let me make that clear.

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So if I can write all of this in terms of sines of theta,

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because the sine of theta is equal to the sine of the arc

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sine of x minus 3 over 2.

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Which is just equal to x minus 3 over 2.

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So if I can write this in terms of sines of theta, then I can

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just make this substitution.

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Sine of theta equals that, and everything

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simplifies a good bit.

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So let's see if we can do that.

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Well, you may or may not know the other identity, and I've

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proven this as well, that sine of 2 theta, that's the same

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thing as sine of theta plus theta, which is equal to sine

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of theta cosine of theta plus, the thetas get swapped around,

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sine of theta plus cosine of theta, which is equal to, this

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is just the same thing written twice, 2 sine of theta

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cosine of theta.

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Some people have this memorized ahead of time, and if you have

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to take an exam on trig substitution, it doesn't hurt

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to have this memorized ahead of time.

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But let's rewrite this like this.

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So our indefinite integral in terms of theta, or our

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antiderivative, became 2 theta, plus, instead of sine of 2

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theta, we could write, 2 sine of theta cosine of theta, and

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of course we have a plus c.

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Now, I want to write everything in terms of sines of theta, but

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I have a cosine if theta there.

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So what can we do?

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Well, we know we know that cosine squared of theta is

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equal to 1 minus sine squared of theta, or that cosine of

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theta is equal to the square root of 1 minus sine

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squared of theta.

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Which seems like we're adding complexity to it, but the neat

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thing is, it's in terms of sine of theta.

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So let's do that.

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Let's make the substitution.

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So our antiderivative, this is the same thing as 2 data plus 2

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sine of theta times cosine of theta, which is equal to this,

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times the square root of 1 minus sine squared of theta,

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and all of that plus c.

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Now we're in the home stretch.

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This problem was probably harder than you thought

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it was going to be.

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We know that sine theta is equal to x minus 3 over 2.

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So let's make that reverse substitution.

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So we have 2 times theta.

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This first term is just 2 times theta, right there.

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So that's 2 times, we can't escape the arc sine.

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If we just have a theta, we have to say that theta is

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equal to arc sine of x minus 3 over 2.

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And then we have plus, let me switch colors,

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plus 2 sine of theta.

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Well, that's plus 2 times sine of theta is x minus 3 over 2.

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So 2 times x minus 3 over 2, and then all of that times

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the square root of 1 minus sine of theta squared.

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What's sine of theta?

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It's x minus 3 over 2 squared.

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And of course we have a plus c.

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Let's see if we can simplify this even more, now that

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we're at the home stretch.

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So this is equal to 2 arc sine of x minus 3 over 2, plus these

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two terms, this 2 and this 2 cancel out, plus x minus 3,

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times the square root of, what happens if we multiply

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everything here by, let's see, if we take a, so this is equal

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to 1 minus x minus 3 over 4. x minus 3 squared over 4.

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This simplification, I should write that in quotes, is taking

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me longer than I thought.

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But let's see if we can simplify this even more.

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If we multiply, let me just focus on this term right here,

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if we multiply the outside, or say, let's multiply

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and divide this by 2.

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So I'll write this as-- so let's just multiply

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this times 2 over 2.

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You might say, Sal, why are you doing that?

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Because I can rewrite this, let me write my whole thing here.

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So I have 2 arc sine of x minus 3 over 2, and then

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I have, I could take this denominator 2 right here.

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So I say, plus x minus 3 over 2.

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That 2 is that 2 right there.

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And then I could write this 2 right here as

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a square root of 4.

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Times the square root of 4, times the square

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root of all of this.

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1 minus x minus 3 squared over 4.

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I think you see where I'm going.

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I'm kind of reversing everything that I did at the

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beginning of this problem.

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And maybe I'm getting a little fixated on making this as

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simple as possible, but I'm so close, so let me finish.

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So I get 2 times the arc sine of x minus 3 over 2, which I'm

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tired of writing, plus x minus 3 over 2, and if we bring this

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4 in, right, the square root of 4 times the square root of that

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is equal to the square root of 4 times these things.

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So it's 4 minus x minus 3 squared, all of

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that that plus c.

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And we're at the home stretch.

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This is equal to 2 times the arc sine of x minus 3 over 2

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plus x minus 3 over 2 times the radical of 4 minus, let's

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expand this, x squared minus 6x plus 9.

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And then this expression right here simplifies to minus minus,

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it's 6x minus x squared, and then you have a 4

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minus 9 minus 5.

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Which is our original antiderivative.

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So finally, we're at the very end.

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We get the antiderivative is 2 arc sine of x minus 3 over 2

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plus x minus 3 over 2 times, time the radical of 6x

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minus x squared minus 5.

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That right there is the antiderivative of what this

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thing that we had at the very top of my little chalkboard,

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which is right there.

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So that is equal to the antiderivative of the

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square root of 6x minus x squared minus 5 dx.

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And I can imagine that you're probably as tired as I am.

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My hand actually hurts.

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But hopefully you find that to be vaguely satisfying.

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Sometimes I get complaints that I only do easy problems.

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Well, this was quite a hairy and not-so-easy problem.