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Let's say we have the indefinite integral of 1

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over 36 plus x squared d x.

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Now, as you can imagine, this is not an easy integral to

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solve without trigonometry.

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I can't do u substitution, I don't have the derivative of

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this thing sitting someplace.

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This would be easy if I had a 2x sitting there.

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Than I would say, oh the derivative of this is 2x,

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I could do u substitution and I'd be set.

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But there is no 2x there, so how do I do it?

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Well, I resort to our trigonometric identities.

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Let's see what trig identity we can get here.

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The first thing I always do, this is just the way my brain

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works, I always like it-- I can see this is a constant plus

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something squared, which tells me I should use a

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trigonometric identity.

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But I always like it in terms of 1 plus something squared.

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I'm just going to rewrite my integral as being equal to,

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let me write the dx in the numerator.

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This is just times dx.

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Let me write a nicer integral than that.

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This is equal to the integral of d x over 36 times 1

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plus x squared over 36.

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1 plus x squared over 36, that's another way to

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write my integral.

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Let's see if any of our trig identities can somehow be

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substituted in here for that that would somehow

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simplify the problem.

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So the one that springs to mind, and if you don't know

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this already, I'll write it down right here, is 1 plus

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tangent squared of theta.

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Let's prove this one.

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Tangent squared of theta, this is equal to 1 plus just the

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definition of tangent sine squared of theta over

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cosine squared of theta.

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Now 1 is just cosine squared over cosine squared.

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So I can rewrite this as equal to cosine squared of theta over

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cosine squared of theta, that's 1, plus sine squared theta over

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cosine squared of theta, now that we have a

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common denominator.

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Now what's cosine squared plus sine squared?

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Definition of the unit circle.

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That equals 1 over cosine squared of theta.

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Or we could say that that equals 1 over cosine squared.

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One over cosine is secant.

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So this is equal to the secant squared of theta.

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If we make the substitution, if we say let's make this thing

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right here equal to tangent of theta, or tangent

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squared of theta.

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Then this expression will be 1 plus tangent squared of theta.

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Which is equal to secant squared.

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Maybe that'll help simplify this equation a bit.

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We're going to say that x squared over 36 is equal to

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tangent squared of theta.

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Let's take the square root of both sides of this equation and

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you get x over 6 is equal to the tangent of theta, or that x

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is equal to 6 tangent of theta.

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If we take the derivative of both sides of this with respect

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to theta we get d x d theta is equal to-- what's the

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derivative of the tangent of theta?

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I could show it to you just by going from these basic

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principles right here.

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Actually let me do it for you just in case.

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So the derivative of tangent theta-- never hurts to do it

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on the side, let me do it right here.

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It's going to be 6 times the derivative with respect to

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theta of tangent of theta.

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Which we need to figure, so let's figure it out.

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The derivative of tangent of theta, that's the same thing

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as d d theta of sine of theta over cosine of theta.

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That's just the derivative of tangent.

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Or this is just the same thing as the derivative with respect

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to theta, let me scroll to the right a little bit.

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Because I never remember the quotient rule, I've told you in

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the past that it's somewhat lame, of sine of theta times

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cosine of theta to the minus 1 power.

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What is this equal to?

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We could say it's equal to, well the derivative of the

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first expression or the first function we could say, which

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is just cosine of theta.

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This is equal to cosine of theta, that's just the

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derivative of sine of theta times our second expression.

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Times cosine of theta to the minus 1.

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I've put these parentheses, and put the minus 1 out there

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because I didn't want to put the minus 1 here and make you

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think that I'm talking about an inverse cosine or an arccosine.

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So that's the derivative of sine times cosine and now I

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want to take plus the derivative of cosine.

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Not just cosine, the derivative if cosine to the minus 1.

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So that is minus 1 times cosine to the minus 2 power of theta.

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That's the derivative of the outside times the

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derivative of the inside.

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Let me scroll over more.

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So that's the derivative of the outside.

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If the cosine theta was just an x, you would say x to the minus

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1 derivative is minus 1 x to the minus 2.

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Now times the derivative of the inside.

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Of cosine of theta with respect to theta.

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So that's times minus sine of theta.

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I'm going to multiply all of that times sine of theta.

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The derivative of this thing, which is the stuff in green,

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times the first expression.

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So what does this equal?

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These cosine of theta divided by cosine of

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theta, that is equal to 1.

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And then I have a minus 1 and I have a minus sine of theta.

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That's plus plus.

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What do I have?

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I have sine squared, sine of theta time sine of theta

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over cosine squared.

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So plus sine squares of theta over cosine squared of theta.

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Which is equal to 1 plus tangent squared of theta.

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What's 1 plus tangent squared of theta?

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I just showed you that.

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That's equal to secant squared of theta.

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So the derivative of tangent of theta is equal to

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secant squared of theta.

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All that work to get us fairly something-- it's nice

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when it comes out simple.

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So d x d theta, this is just equal to secant

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squared of theta.

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If we want to figure out what d x is equal to, d x is equal to

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just both sides times d theta.

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So it's 6 times secant squared theta d theta.

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That's our d x.

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Of course, in the future we're going to have to back

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substitute, so we want to solve for theta.

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That's fairly straightforward.

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Just take the arctangent of both sides of this equation.

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You get that the arctangent of x over 6 is equal to the theta.

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We'll save this for later.

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So what is our integral reduced to?

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Our integral now becomes the integral of d x?

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What's d x?

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It is 6 secant squared theta d theta.

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All of that over this denominator, which is 36

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times 1 plus tangent squared of theta.

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We know that this right there is secant squared of theta.

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I've shown you that multiple times.

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So this is secant squared of theta in the denominator.

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We have a secant squared on the numerator, they cancel out.

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So those cancel out.

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So are integral reduces to, lucky for us, 6/36 which

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is just 1/6 d theta.

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Which is equal to 1/6 theta plus c.

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Now we back substitute using this result.

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Theta is equal to arctangent x over 6.

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The anti-derivative 1 over 36 plus x squared is

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equal to 1/6 times theta.

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Theta's just equal to the arctangent x over 6 plus c.

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And we're done.

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So that one wasn't too bad.

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