WEBVTT mathematics/calculus-ii/murray
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Welcome to educator.com, this is the lecture on trigonometric substitution.
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There are three main equations that you have for trigonometric substitution.
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The idea is that if you see any one of these three forms in an integral, then you can do what is called a trigonometric substitution to convert your integral into a trigonometric integral.
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Then hopefully you can use some of the trigonometric techniques that we learned in the previous lecture to solve the integral.
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Let me show you how they work.
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If you see something like the sqrt(a-bu²).
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Now here the a and the b are constants, and the u is the variable.
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You are going to make the substitution u=sqrt(a/(bsin(θ)).
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Why does that work?
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Well, if you do that, then u² is a/(bsin²θ),
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And, so a-bu² will be a - while bu² will be asin²(θ)
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That is a × (1-sin²(θ))
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And 1-sin²(θ) of course is cos²(θ).
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If you have the square root of that expression, then that will convert into cos(θ), and you will get an easier integral to deal with.
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By the same idea, if you have au²-b, then you are going to have u=sqrt(b/(asec(θ)).
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What you are trying to do there is take advantage of the trigonometric identity tan²(θ)+1=sec²(θ).
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Sec²(θ)-1 = tan²(θ) so once you make this substitution you are going to end up with sec²(θ)-1 under the square root.
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The last one you will have a+bu²=sqrt(a/(btan(θ))) and so again you are taking advantage of this trigonometric identity tan²(θ)+1.
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You will end up with that under the square root, and that will convert into sec²(θ).
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All of these take advantage of these trigonometric identities.
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One thing that is important to remember when you are making these substitutions is whenever you substitute u equals something, or x equals something, you always have to substitute in dU or dX as well.
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For example, if you substitute u=sqrt(a/bsin(θ)), then you also have to substitute dU, which would be, a and b are constants so that is just sqrt(a/bcos(θ)) dθ.
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You always have to make the accompanying substitution for your dU or your dX.
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As with all mathematical problems it is a little hard to understand when you are just looking at mathematical formulas in general, but we will move onto examples and you will see how these work.
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The first example is the integral of 4 - 9x² dx.
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That example matches the first pattern that we saw before, which was the square root of a minus b u².
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The substitution that we learned for that is u equals the square root of a/b sin(θ).
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Here, the a is 4, the b is 9, the x is taking the place of the u.
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What we are going to substitute is x equals well the square root of a over b, is the square root of 4/9 sin(θ).
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The square root of 4/9 is 2/3 sin(θ).
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As we do that we also remember we have to substitute dX, which is going to be 2/3 cos(θ) dθ.
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This is why it is really important to write the dX along with your integral.
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It helps you remember that you also need to do the extra part of the substitution with the dX.
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You have to convert that to the new variable as well.
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So, then our integral becomes the integral of the square root of 4 minus, well 9x² is 9, x² is 4/9 sin²(θ) and dX is 2/3 cos(θ) dθ.
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Just working inside the square root for a moment here, we get the square root of 4 minus 4 sin²(θ) and we can pull the 4 out of the square root.
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We get 4 times the square root of 1 minus sin²θ.
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That is 2 times 1 minus sin²(θ) is cos²(θ) so that will give us cos(θ).
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Now, if we bring in the other elements of the integral, we get the integral.
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I am going to collect all of the constants outside, so we have a 2, and a 2/3 that is 4/3 cos²(θ) because we have one cos here, and one cos here, dθ.
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We learned how to integrate cos²(θ), you use the half angle formula.
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This is 4/3 times the integral of 1/2 times one plus cos(2θ) dθ.
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If we combine the 1/2 with the 4 we get 2/3 times now the integral of 1.
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Remember we're integrating with respect to θ so that is θ plus the integral of cos(2θ).
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Well the integral of cos is sin, but because of the 2 there, we have to put a 1/2 there.
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This is now 2/3 and we want to convert this back to x's.
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We have to solve these equations for θ in terms of x.
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θ if we solve this equation in terms of x, we get 3/2 x equals sin(θ)
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So θ is equal to arcsin(3/2 x)
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Arcsin(3/2 x), that is the θ. Now 1/2 sin(2θ) to sort that out it helps to remember that sin(2θ) is 2 times sin(θ) times cos(θ)
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1/2 2sin(θ) is sin(θ) times cos(v)
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So, 2/3 arcsin(3/2 x)
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Now, sin(θ) we said was 3/2 x.
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Cos(θ) is the square root of 1 minus sin²(θ), so 1 minus 3/2 sin(θ) was x 9/4 x².
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Now we have converted everything back in terms of x.
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And we have an answer.
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The point here was that we started with a square root of a quadratic and we used our trigonometric substitution and we converted this integral into a trigonometric integral.
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Then we learned in the other lecture how to convert trigonometric integrals, so we solved that integral in terms of θ.
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Then we have to convert it back into terms of x.
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So, let us try another example, this one is actually a little quicker.
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We see the integral of dX over 1 + x²
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The key thing to remember here is that when you see the square root of 1 + x², or even 1 + x² without a root, you want to use the tangent substitution.
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X equals the tan(θ)
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The reason you do that is that x² + 1 is tangent²(θ) + 1 but that is sec²(θ)
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Let us make that substitution x equals tanθ.
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Of course, every time you make a substitution you also have to make a substitution for dX.
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dX is equal to sec²(θ) dθ
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If we plug that into the integral, we get dX is sec²(θ) dθ.
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1 + x² is sec²(θ)
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We kind of lucked out on this one because the sec²'s cancel and we just get the integral of θ
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This turns out to be a really easy one, that is just θ.
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θ, remember that x was tan(θ) we know that θ is arctan(x) + a constant.
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The thing to remember about this example is that when you have a + there, you are going to go for a tangent substitution.
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Whenever you have a minus, you are going to go for a secant substitution.
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We are going to move to a more complicated example now.
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dX/x² + 8x + 25.
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We have a more complicated expression in the denominator here.
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What we are going to do is try to simplify it into something that admits a trigonometric substitution pretty easily.
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What we are going to do is a little high school algebra on the denominator.
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We are going to complete the square on that, so that is x²
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Remember, the way you complete the square is you take this 8 and you divide by 2 and you square it.
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So 8 divide by 2 is 4 and 4 squared is 16, so we will write 16 there.
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To make this a true equation, we have to add 9 there.
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Then that is x + 4, quantity squared + 9.
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We are going to use that for the integral.
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What we are going to do is use that for the quick substitution.
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U is equal to x + 4 and again we have to convert dU, but that is easy that is just dX.
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The integral just converts into dU/u² + 9.
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Now, to finish that integral, well, we want to make a trigonometric substitution.
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We are going to let u be 3tanθ.
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The way I got 3tanθ was I looked at 9 and I took the square root of that and I got 3.
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The point of doing that is that u² + 9 will be 9tan²(θ)+9.
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And, that is 9+tan²(θ)+1.
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Which is 9 sec²(θ).
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Of course with any substitution, you have to figure out what dU is.
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Well if u is 3tan(θ) then dU is 3 sec²(θ) dθ.
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We will plug all of that in, the dU was 3 sec²θ dθ.
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The u² + 9 converted into 9 sec²θ
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The three and the 9 simplify down into 1/3 and now again, the secants cancel and we just have the integral of dθ.
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That is 1/3 θ.
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θ is something we can figure out from this equation over here.
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θ if we solve this equation u/3=tanθ.
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θ is arctan(u/3).
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But we are not finished with that because we still have to convert back into terms of x.
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That is 1/3 arctan, now u was (x+4), remember our original substitution there, so (x+4)/3 and now we add on a constant.
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Now we are done.
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OK, that is the end of the first instalment of trigonometric substitution.
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This has been educator.com.