College Calculus: Level II Trigonometric Substitutions

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Here are a couple more examples on trigonometric substitution.0003

We are going to start with the integral of 6x x² dx.0006

We have a quadratic under the sqrt.0012

The problem is we do not have a constant - x².0016

We have 6x - x².0020

What we have to do is complete the square on that before we go ahead with our trigonometric substitution.0022

Because we do not have a constant yet.0030

I am going to write that as sqrt, I am going to factor out the negative sign first, and I will get - x² - 6x,0031

And now I want to complete the square here.0043

So, I take the middle term, the 6x, and divide it by 2 and square it.0046

So -6/-2 is 3, you square that and you get 9, add on 9.0051

Now to pay for that, well I added 9, but that was inside the negative sign right here.0059

So, what I really did there was I subtracted 9,0069

So, to balance that out I have to add 9.0072

We get sqrt(-x²-6x+9)+9.0079

I will write that as sqrt(9) -, and the whole point of completing the square there is that this was (x-3)².0090

The first thing I am going to do here is a little substitution.0102

Let u = (x-3), and we always have to substitute the du as well.0109

The du is dx there, so it is an easy substitution but we have to do it.0112

What we have here is the integral of sqrt(9-u²) du.0120

Now, this is something that looks like it is ready for a trig substitution.0131

We look at the trig substitution rules and we see that this one calls for a sin substitution,0135

Because it is 9 minus u², the u is being subtracted.0142

So we are going to let u = 3sin(θ).0146

Again, I got this 3 because it is the square root of this nine.0154

du is going to be 3cos(θ) dθ.0160

The point of doing that is the sqrt(9-u²), well, u² = 9²θ.0171

If we pull a 9 out of the radical we get 3 × (1-sin²θ).0190

1-sin²θ is cos²θ, so this is just 3cos(θ).0199

The radical turns into 3cosθ, the du turns into 3cos(θ) dθ.0204

This whole integral turns into 9cos²θ dθ.0219

Again, we have a trigonometric integral in the trigonometric integral lecture, that when you see even powers of cosine,0228

you use the half-angle formula.0233

This is 9 × the integral of 1/2.0237

I will just pull the 1/2 outside, 9/2, times the integral of 1 + cos(2θ) dθ.0243

So, that is 9/2, if you integrate 1 you get θ + integral of cos is sine,0256

But because of that 2, we have to have a 1/2 there, and this was all being multiplied by the 9/2.0272

We will still have a constant.0282

Again, we have sin(2θ).0285

We have seen this earlier and the way we want to resolve sin(2θ) is 2sin(θ) cos(θ).0290

So, I am going to keep going on the next page here.0301

Let us remember the important things are that u was 3sin(θ), and in turn u was (x-3).0307

9 - u² was 3cos(θ).0316

We will be using the sin(2θ) and I am just going to copy this equation on the next page and keep going.0322

From the previous page we have 9/2 × θ + 1/2 sin(2θ) + C.0331

So, 9/2, how can we sort θ out?0347

Well remember that u was 3sinθ.0353

So, θ if you solve this, u/3 is sin(θ).0364

θ is arcsin(u/3).0372

The 1/2 sin(2θ), remember sin(2θ) is 2sin(θ) cos(θ).0378

So 1/2 sin(2θ) is sin(2θ) cos(θ).0390

So we have 9/2 arcsin(u/3), sin(θ) is u/3.0403

Now the cos(θ), what we remember from before is that cos(θ).0416

Actually we cannot see it here, so maybe I will work it out again very quickly.0424

Cos(θ) is the sqrt(1 - sin²(θ)), which is sqrt(1) -, sin(θ) is u/3, so we know sin(θ) is u²/90430

Cos(θ) is sqrt(1 - u²/9).0445

And, so, this turns into 9/2 arcsin, OK u/3 it is time to convert that back into, remember u is x-3.0456

So x-3/3 + now we have u/3 again.0470

That is x-3/3 again.0475

Now 1 - u²/9 is, 1 - u² is (x-3)²/9.0483

I am just going to try to simplify this radical because I think you will recognize it after I simplify it.0498

That is 1 - (x-3)², is x² - 6x + 9/9.0504

We are going to write 1 as 9/9.0518

The 9/9's cancel, and we get 9/2 arcsin(x-3/3) + (x-3)/3.0524

This radical turns into over 9 - x² + 6x.0538

If you pull that 9 out, we get a 1/3 6x-x², which is the same radical we started with.0547

So, our final answer is 9/2 arcsin(x-3/3) + x-3/9 × sqrt(6x - x²) + C.0559

That one was a pretty long example.0581

The key to it was recognizing this initial radical in the integral.0589

Recognizing that we had to complete the square on that.0594

Then we had to do a trigonometric sin substitution on that.0600

So, those were sort of the two key theoretical steps there.0611

The rest was just keeping track of lots of substitutions.0617

X into u, and u into θ0618

And then keeping track of a lot of different constants that were percolating throughout the whole interval.0625

A very long and complicated example but the basic ideas were just completing the square and doing a basic substitution.0630

So I want to do one more example.0000

This is the integral of dx/sqrt(x² + 1).0004

Remember, when you see x² + 1, that is your sort of warning flag that you are going to need a tangent substitution.0009

Remember a minus tells you that you are going to use either a sin or a secant substitution,0018

And a plus tells us that we are going to use a tangent substitution.0025

So we are going to use x = tan(θ), and then dx = sec²(θ) dθ.0027

x² + 1 = tan²(θ + 1), which by the trigonometric identity is sec²(θ).0043

And, that was the whole point of making the substitution, to invoke that trigonometric identity,0059

That we would get tan² + 1 and could convert it into sec²(θ).0064

The sqrt(sec²(θ) is just sec(θ).0072

When we solve this integral, or rather make this substitution into this integral,0077

We have dx in the numerator, so that converts into sec²(θ) dθ.0087

In the denominator we have the sqrt(x² + 1),0093

We already saw that that is sec(θ).0094

So, the secants, one of the secants cancels and we just get the integral of sec(θ) dθ.0100

That is the trigonometric integral that we learned in the trigonometric integral lecture.0113

You have to remember how to do it.0116

There is an old trick, where you multiply the top and bottom by sec(θ) tan(θ).0121

We covered that in the lecture on trigonometric integrals.0136

The answer for the integral of sec(θ) came out to be the ln(abs(sec(θ) + tan(θ).0140

That in turn, turns into, remember we have to substitute back into x.0153

sec(θ), I figure out what this right here, that is the sqrt of x² + 1.0156

tan(θ) was our original substitution, that was x.0164

And, we have to put a constant on that.0168

So, that is our answer.0175

Again, the key step there was recognizing that we had x² + 1,0177

And recognizing that that would give us a tangent substitution.0181

It was also important to keep track of the dx,0188

And then to work everything into θ's, and then at the end convert everything back into x's.0190

So that is the end of the lecture on trigonometric substitutions.0195

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Welcome to educator.com, this is the lecture on trigonometric substitution.0000

There are three main equations that you have for trigonometric substitution.0006

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.0012

Then hopefully you can use some of the trigonometric techniques that we learned in the previous lecture to solve the integral.0028

Let me show you how they work.0035

If you see something like the sqrt(a-bu²).0038

Now here the a and the b are constants, and the u is the variable.0042

You are going to make the substitution u=sqrt(a/(bsin(θ)).0048

Why does that work?0051

Well, if you do that, then u² is a/(bsin²θ),0053

And, so a-bu² will be a - while bu² will be asin²(θ)0063

That is a × (1-sin²(θ))0079

And 1-sin²(θ) of course is cos²(θ).0088

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.0096

By the same idea, if you have au²-b, then you are going to have u=sqrt(b/(asec(θ)).0104

What you are trying to do there is take advantage of the trigonometric identity tan²(θ)+1=sec²(θ).0117

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.0131

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)+1.0144

You will end up with that under the square root, and that will convert into sec²(θ).0161

All of these take advantage of these trigonometric identities.0165

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.0170

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.0182

You always have to make the accompanying substitution for your dU or your dX.0204

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.0211

The first example is the integral of 4 - 9x² dx.0223

That example matches the first pattern that we saw before, which was the square root of a minus b u².0229

The substitution that we learned for that is u equals the square root of a/b sin(θ).0237

Here, the a is 4, the b is 9, the x is taking the place of the u.0249

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).0259

The square root of 4/9 is 2/3 sin(θ).0276

As we do that we also remember we have to substitute dX, which is going to be 2/3 cos(θ) dθ.0279

This is why it is really important to write the dX along with your integral.0291

It helps you remember that you also need to do the extra part of the substitution with the dX.0296

You have to convert that to the new variable as well.0302

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.0304

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.0334

We get 4 times the square root of 1 minus sin²θ.0343

That is 2 times 1 minus sin²(θ) is cos²(θ) so that will give us cos(θ).0352

Now, if we bring in the other elements of the integral, we get the integral.0360

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.0363

We learned how to integrate cos²(θ), you use the half angle formula.0380

This is 4/3 times the integral of 1/2 times one plus cos(2θ) dθ.0387

If we combine the 1/2 with the 4 we get 2/3 times now the integral of 1.0401

Remember we're integrating with respect to θ so that is θ plus the integral of cos(2θ).0412

Well the integral of cos is sin, but because of the 2 there, we have to put a 1/2 there.0421

This is now 2/3 and we want to convert this back to x's.0432

We have to solve these equations for θ in terms of x.0448

θ if we solve this equation in terms of x, we get 3/2 x equals sin(θ)0449

So θ is equal to arcsin(3/2 x)0465

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)0478

1/2 2sin(θ) is sin(θ) times cos(v)0500

So, 2/3 arcsin(3/2 x)0515

Now, sin(θ) we said was 3/2 x.0525

Cos(θ) is the square root of 1 minus sin²(θ), so 1 minus 3/2 sin(θ) was x 9/4 x².0532

Now we have converted everything back in terms of x.0554

And we have an answer.0562

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.0568

Then we learned in the other lecture how to convert trigonometric integrals, so we solved that integral in terms of θ.0588

Then we have to convert it back into terms of x.0592

So, let us try another example, this one is actually a little quicker.0598

We see the integral of dX over 1 + x²0605

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.0611

X equals the tan(θ)0629

The reason you do that is that x² + 1 is tangent²(θ) + 1 but that is sec²(θ)0632

Let us make that substitution x equals tanθ.0642

Of course, every time you make a substitution you also have to make a substitution for dX.0650

dX is equal to sec²(θ) dθ0653

If we plug that into the integral, we get dX is sec²(θ) dθ.0663

1 + x² is sec²(θ)0672

We kind of lucked out on this one because the sec²'s cancel and we just get the integral of θ0680

This turns out to be a really easy one, that is just θ.0686

Theta, remember that x was tan(θ) we know that θ is arctan(x) + a constant.0692

The thing to remember about this example is that when you have a + there, you are going to go for a tangent substitution.0706

Whenever you have a minus, you are going to go for a secant substitution.0720

We are going to move to a more complicated example now.0727

dX/x² + 8x + 25.0729

We have a more complicated expression in the denominator here.0737

What we are going to do is try to simplify it into something that admits a trigonometric substitution pretty easily.0738

What we are going to do is a little high school algebra on the denominator.0748

We are going to complete the square on that, so that is x²0758

Remember, the way you complete the square is you take this 8 and you divide by 2 and you square it.0764

So 8 divide by 2 is 4 and 4 squared is 16, so we will write 16 there.0770

To make this a true equation, we have to add 9 there.0777

Then that is x + 4, quantity squared + 9.0780

We are going to use that for the integral.0791

What we are going to do is use that for the quick substitution.0795

U is equal to x + 4 and again we have to convert dU, but that is easy that is just dX.0797

The integral just converts into dU/u² + 9.0811

Now, to finish that integral, well, we want to make a trigonometric substitution.0820

We are going to let u be 3tanθ.0829

The way I got 3tanθ was I looked at 9 and I took the square root of that and I got 3.0839

The point of doing that is that u² + 9 will be 9tan²(θ)+9.0844

And, that is 9+tan²(θ)+1.0856

Which is 9 sec²(θ).0863

Of course with any substitution, you have to figure out what dU is.0867

Well if u is 3tan(θ) then dU is 3 sec²(θ) dθ.0871

We will plug all of that in, the dU was 3 sec²θ dθ.0881

The u² + 9 converted into 9 sec²θ0890

The three and the 9 simplify down into 1/3 and now again, the secants cancel and we just have the integral of dθ.0900

That is 1/3 θ.0907

θ is something we can figure out from this equation over here.0914

θ if we solve this equation u/3=tanθ.0925

θ is arctan(u/3).0933

But we are not finished with that because we still have to convert back into terms of x.0938

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.0944

Now we are done.0960

OK, that is the end of the first instalment of trigonometric substitution.0966

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