For more information, please see full course syllabus of College Calculus: Level II

For more information, please see full course syllabus of College Calculus: Level II

### Trigonometric Substitutions

**Main formula:**

For , substitute .

For , substitute .

For , substitute .

**Hints and tips:**

Remember to make the substitution as well as the main substitution.

You can sometimes use trigonometric substitution even when you don’t have a square root. However, in these cases, you should only be using the

__tangent__substitution. If you get one of the other ones, then you should have been able to factor the algebraic expression, and trigonometric substitution was not necessary.If you have a linear term (

*x*or*u*, as opposed to*x*² or*u*²), complete the square and make a substitution to eliminate the linear term before you make the trigonometric substitution.If the coefficient of the quadratic term (

*x*²) is negative, factor the negative out of all terms before completing the square.After you make the trigonometric substitution, it becomes a trigonometric integral that you can handle using the techniques in the previous section.

### Trigonometric Substitutions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Important Equations 0:06
- How They Work
- Example
- Remember: du and dx
- Lecture Example 1 3:43
- Lecture Example 2 10:01
- Lecture Example 3 12:04
- Additional Example 4
- Additional Example 5

### College Calculus 2 Online Course

I. Advanced Integration Techniques | ||
---|---|---|

Integration by Parts | 24:52 | |

Integration of Trigonometric Functions | 25:30 | |

Trigonometric Substitutions | 30:09 | |

Partial Fractions | 41:22 | |

Integration Tables | 20:00 | |

Trapezoidal Rule, Midpoint Rule, Left/Right Endpoint Rule | 22:36 | |

Simpson's Rule | 21:08 | |

Improper Integration | 44:18 | |

II. Applications of Integrals, part 2 | ||

Arclength | 23:20 | |

Surface Area of Revolution | 28:53 | |

Hydrostatic Pressure | 24:37 | |

Center of Mass | 25:39 | |

III. Parametric Functions | ||

Parametric Curves | 22:26 | |

Polar Coordinates | 30:59 | |

IV. Sequences and Series | ||

Sequences | 31:13 | |

Series | 31:46 | |

Integral Test | 23:26 | |

Comparison Test | 22:44 | |

Alternating Series | 25:26 | |

Ratio Test and Root Test | 33:27 | |

Power Series | 38:36 | |

V. Taylor and Maclaurin Series | ||

Taylor Series and Maclaurin Series | 30:18 | |

Taylor Polynomial Applications | 50:50 |

### Transcription: Trigonometric Substitutions

*This is educator.com.*0000

*Here are a couple more examples on trigonometric substitution.*0003

*We are going to start with the integral of 6x x ^{2} dx.*0006

*We have a quadratic under the sqrt.*0012

*The problem is we do not have a constant - x ^{2}.*0016

*We have 6x - x ^{2}.*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 - x ^{2} - 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(-x ^{2}-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-u ^{2}) 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 u ^{2}, 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-u ^{2}), well, u^{2} = 9^{2}θ.*0171

*If we pull a 9 out of the radical we get 3 × (1-sin ^{2}θ).*0190

*1-sin ^{2}θ is cos^{2}θ, 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 9cos ^{2}θ 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 - u ^{2} 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 - sin ^{2}(θ)), which is sqrt(1) -, sin(θ) is u/3, so we know sin(θ) is u^{2}/9*0430

*Cos(θ) is sqrt(1 - u ^{2}/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 - u ^{2}/9 is, 1 - u^{2} 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 x^{2} - 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 - x ^{2} + 6x.*0538

*If you pull that 9 out, we get a 1/3 6x-x ^{2}, 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 - x ^{2}) + 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 θ*0618

*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(x ^{2} + 1).*0004

*Remember, when you see x ^{2} + 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 = sec ^{2}(θ) dθ.*0027

*x ^{2} + 1 = tan^{2}(θ + 1), which by the trigonometric identity is sec^{2}(θ). *0043

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

*That we would get tan ^{2} + 1 and could convert it into sec^{2}(θ).*0064

*The sqrt(sec ^{2}(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 sec ^{2}(θ) dθ.*0087

*In the denominator we have the sqrt(x ^{2} + 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 x ^{2} + 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 x ^{2} + 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

*This has been educator.com.*0200

*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-bu ^{2}).*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 u ^{2} is a/(bsin^{2}θ), *0053

*And, so a-bu ^{2} will be a - while bu^{2} will be asin^{2}(θ)*0063

*That is a × (1-sin ^{2}(θ))*0079

*And 1-sin ^{2}(θ) of course is cos^{2}(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 au ^{2}-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 tan ^{2}(θ)+1=sec^{2}(θ).*0117

*Sec ^{2}(θ)-1 = tan^{2}(θ) so once you make this substitution you are going to end up with sec^{2}(θ)-1 under the square root.*0131

*The last one you will have a+bu ^{2}=sqrt(a/(btan(θ))) and so again you are taking advantage of this trigonometric identity tan^{2}(θ)+1.*0144

*You will end up with that under the square root, and that will convert into sec ^{2}(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 - 9x ^{2} dx.*0223

*That example matches the first pattern that we saw before, which was the square root of a minus b u ^{2}.*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 9x ^{2} is 9, x^{2} is 4/9 sin^{2}(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 sin ^{2}(θ) and we can pull the 4 out of the square root.*0334

*We get 4 times the square root of 1 minus sin ^{2}θ.*0343

*That is 2 times 1 minus sin ^{2}(θ) is cos^{2}(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 cos ^{2}(θ) because we have one cos here, and one cos here, dθ.*0363

*We learned how to integrate cos ^{2}(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 sin ^{2}(θ), so 1 minus 3/2 sin(θ) was x 9/4 x^{2}.*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 + x ^{2} *0605

*The key thing to remember here is that when you see the square root of 1 + x ^{2}, or even 1 + x^{2} without a root, you want to use the tangent substitution.*0611

*X equals the tan(θ)*0629

*The reason you do that is that x ^{2} + 1 is tangent^{2}(θ) + 1 but that is sec^{2}(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 sec ^{2}(θ) dθ*0653

*If we plug that into the integral, we get dX is sec ^{2}(θ) dθ.*0663

*1 + x ^{2} is sec^{2}(θ) *0672

*We kind of lucked out on this one because the sec ^{2}'s cancel and we just get the integral of θ*0680

*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/x ^{2} + 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 x ^{2} *0758

*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/u ^{2} + 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 u ^{2} + 9 will be 9tan^{2}(θ)+9.*0844

*And, that is 9+tan ^{2}(θ)+1.*0856

*Which is 9 sec ^{2}(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 sec ^{2}(θ) dθ.*0871

*We will plug all of that in, the dU was 3 sec ^{2}θ dθ.*0881

*The u ^{2} + 9 converted into 9 sec^{2}θ*0890

*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

*This has been educator.com.*0972

1 answer

Last reply by: Dr. William Murray

Thu Aug 4, 2016 6:01 PM

Post by Peter Ke on July 27, 2016

For example 3, you let u = 3tan(theta). Where did that idea come from and why?

1 answer

Last reply by: Dr. William Murray

Sat Feb 27, 2016 10:36 AM

Post by Mohsin Alibrahim on February 25, 2016

Dr Murray,

In example 3, how come du = [3sec^2(x)] while u = 9sec^2(x) ?

Thanks for the wonderful lecture

3 answers

Last reply by: Dr. William Murray

Fri Apr 10, 2015 1:07 PM

Post by Luvivia Chang on April 7, 2015

Hello Dr. Murray, I have a question about exemple 1

i think the square root of (1-sin^2(theta))=the square root of cos^2(theta)= the "absolute value of cos(theta)" instead of just cos (theta)

Am I correct? Thanks

2 answers

Last reply by: Dr. William Murray

Thu Feb 19, 2015 3:48 PM

Post by Thadeus McNamara on February 18, 2015

Dr. Murray, I am taking the Calc BC Exam. Are there any lectures in this College Calculus II playlist that I can skip because they are not a part of the exam? Thanks

2 answers

Last reply by: Dr. William Murray

Thu Feb 19, 2015 3:48 PM

Post by Thadeus McNamara on February 18, 2015

@4:51, isn't the integral of 2/3sintheta equal to -2/3costheta? You put the integral as 2/3costheta.

3 answers

Last reply by: Dr. William Murray

Wed Aug 27, 2014 9:46 AM

Post by brody wagerson on August 24, 2014

I believe there is an error in lecture example 4, on the last step to simplify the answer you cannot divide x^2-6x+9 by 9 to get x^2-6x, because they are addition and not multiplication. You must multiply out what is under the square root sign to get 9-(x^2-6x+9) the 9's will then cancel and give you 6x-x^2 under the square root just like in the beginning of the question, but without taking out an extra 1/3 to divide. Sorry if my explanation does not make sense I wrote this quickly. but i believe the correct answer to example 4 if I am correct will be 9/2[arcsin(x-3/3)+(x-3/3)*sqrt(6x-x^2)]+C. No stranger to criticism though so let me know if I made a mistake and professor Murray is right.

1 answer

Last reply by: Dr. William Murray

Tue Aug 5, 2014 3:18 PM

Post by Pedro Valdericeda on July 21, 2014

Thank you Prof. Murray. You are really good.

1 answer

Last reply by: Dr. William Murray

Tue Jun 17, 2014 12:03 PM

Post by ahmed ahmed on June 9, 2014

thank you professor! you are awesome!

1 answer

Last reply by: Dr. William Murray

Mon Mar 17, 2014 1:13 PM

Post by Peidong, He on March 16, 2014

WANNA SAY THANK YOU!

1 answer

Last reply by: Dr. William Murray

Fri Aug 23, 2013 11:52 AM

Post by Charles Zhou on August 21, 2013

This is absolutely a nice video! It does help me a lot. I just have a question. I'm a high school student and I wonder is there any difference or relationship between AP calculus BC and this College Calculus Level II

1 answer

Last reply by: Dr. William Murray

Thu Dec 6, 2012 5:40 PM

Post by Samuel Bass on December 6, 2012

In problem #1 in the solution.

You converted cos(theta) = squar(1-sin^2(theta))

yes this would be correct if the cos(theta) was squared e.g. cos^2(theta)

but in this case the cos(theta) is not raised to the power 2.... then you can't use the pythagoream Identitie as you did in this case..... Im I correct? Cause I can't see this specific step you did possible. Please Explain.

1 answer

Last reply by: Dr. William Murray

Mon Oct 22, 2012 1:48 PM

Post by Keisha Baxter on October 21, 2012

He factored incorrectly in Lecture example 1... in about the third line when he factors out the 4 from sqrt(4 - 4 sin^2 x) he puts 2 sqrt (1 - sin^2 x). So all of the coefficients following that is incorrect

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 12:24 PM

Post by SOUFIANE LAMOUNI on May 16, 2012

you are a great Professor , thank you

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 12:23 PM

Post by Kumar Sandrasegaran on February 26, 2012

Very very interesting. Makes difficult calculations look simple

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 12:20 PM

Post by Real Schiran on November 22, 2011

Very Nice, Thanks...........

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 12:19 PM

Post by Jaspreet Singh on February 21, 2010

this is so much more helpful than school! thanks for being so clear

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 12:19 PM

Post by lynette stevenson on February 17, 2010

excellent lecture