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

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Integration of Trigonometric Functions

**Main formula:**

**Hints and tips:**

Remember to check which of the powers is

__odd__. Then let*u*be the__other__one. (E.g. if you have cos^{5}*x*, then use*u*= sin*x*.)If both powers are even, use the half-angle formulas.

If you have an

__even__power of secant, use*u*= tan*x*. If you have an__odd__power of tangent, use*u*= sec*x*.If you have an odd power of secant and an even power of tangent, then follow this process:

Convert tangents to secants, two at a time, using tan² +1 = sec² . You’ll end up with

__odd__powers of secant.Use integration by parts twice (

*u*= sec^{n−2}*x*,*dv*= sec²*x dx*) to cut down from ∫ sec^{n}*x dx*to ∫ sec^{n−2}*x dx*, so you eventually get back to ∫ sec*x dx*. Then remember the formula:

### Integration of Trigonometric Functions

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 Equation 0:07
- Powers (Odd and Even)
- What To Do
- Lecture Example 1 1:37
- Lecture Example 2 3:12
- Half-Angle Formulas 6:16
- Both Powers Even
- Lecture Example 3 7:06
- Lecture Example 4 10:59
- Additional Example 5
- Additional Example 6

### 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: Integration of Trigonometric Functions

*OK, lset us try a new example.*0000

*We have here tan ^{3}(x) × sec(x) and remember that the trick here is to look at the powers.*0004

*You first look at the powers of tan(x).*0009

*Is that even?*0012

*Well 3 is odd so that does not work so we cannot use our strategy there.*0014

*The power on sec(x) is 1.*0020

*We ask, is that an odd number?*0023

*That does work, so we are going to use u = sec(x).*0026

*Then du = sec(x) tan(x) dx.*0033

*The point of that is that we can then convert this integral into something just involving sec(x).*0043

*We have tan ^{3}(x), so we will write that as tan^{2}(x) × tan(x) × sec(x) dx.*0051

*Now the u is going to be sec(x), but we still need to have du.*0064

*sec(x) tan(x).*0066

*We have that right here, sec(x) tan(x).*0069

*The tan ^{2}(x), we need to convert that into secants.*0073

*Since we have an even number of tangents, we can do that.*0078

*Remember our old Pythagorean Identity, tan ^{2}(x) + 1 = sec^{2}(x).*0081

*The tan ^{2}(x) converts into sec^{2}(x) - 1.*0090

*tan(x) sec(x) dx.*0099

*Now we are set up to put our substitution in.*0103

*Sec ^{2}(x) - 1 is u^{2} - 1.*0106

*tan(x) sec(x) dx, that is just du.*0112

*All of a sudden we have a really easy integral.*0117

*We can integrate that quickly to be u ^{3}/3 - u.*0121

*Then we substitute back into sec ^{3}(x)/3 - sec(x) + C.*0126

*So, again, the reason that worked is because we were checking the odds and even powers on the tangent and the secant.*0143

*It worked because we had an odd power of sec(x).*0155

*There is one more example I want to work through with you.*0000

*It is a pretty tricky one.*0003

*It is sec ^{3}(x).*0005

*This one is not so obvious.*0006

*What we are going to do is let u = sec(x).*0010

*But the point of that is not to make a substitution.*0017

*The point of that is to use integration by parts.*0022

*If u = sec(x), then dv = sec ^{2}(x) dx.*0024

*That is the stuff left over after you use one of these secants.*0032

*That is not obvious that we should do it that way.*0036

*You might ask how did you know to break it up that way?*0041

*Well, here is the idea.*0044

*I know that sec ^{2} is the derivative of tangent.*0046

*So, by making dv be sec ^{2} v becomes tan(x).*0050

*That is a relatively simple integration.*0056

*By making u = sec(x), we have du = sec(x) tan(x) dx.*0061

*That looks a little more complicated.*0070

*We will see how it works out.*0074

*Remember the formula for integration by parts.*0075

*It is always uv - the integral of vdu.*0077

*So, uv is sec(x) tan(x) - the integral of vdu.*0081

*So vdu looks a little complicated.*0086

*We have sec(x) tan(x) dx × tan(x).*0096

*So sec(x) × tan ^{2}(x) dx.*0102

*This is looking a little complicated.*0108

*What I am going to do is take this tan(x) and remember the Pythagorean Identity.*0114

*tan ^{2}(x) + 1 is sec^{2}(x).*0116

*I am going to write that tan(x) as sec ^{2}(x) - 1.*0123

*The whole thing becomes sec(x) tan(x) - now I will distribute this secant across these two terms.*0135

*So we have - the integral of sec ^{3}(x) dx.*0147

*+ the integral of sec(x) dx.*0156

*Now, it looks like we have taken what seemed like a pretty hard integral and replaced it by another hard integral.*0163

*And, an integral that is exactly the same as what we started with.*0171

*It looks like it is getting worse and worse.*0175

*Let me show you how you can resolve those two.*0178

*First of all, the sec(x) dx, there is a trick here.*0181

*Which is to multiply by sec(x) + tan(x)/sec(x) + tan(x).*0184

*You are multiplying by 1, not changing it.*0204

*But you get something very interesting when you multiply that into the integral.*0209

*The numerator will be sec(x) × (sec(x) + tan(x)).*0213

*So, sec ^{2}(x) + sec(x) tan(x).*0220

*The denominator is going to be sec(x) + tan(x) all multiplied by dx.*0224

*Look at this.*0238

*If you look at the denominator, sec(x) + tan(x), the derivative of sec(x) is sec(x) tan(x).*0240

*The derivative of tan(x) is sec ^{2}(x).*0250

*So the numerator is the derivative of the denominator.*0255

*So this last integral here is a derivative divided by the original function.*0259

*That last integral just integrates to ln(sec(x)) + tan(x).*0266

*That is a pretty special trick that is probably worth memorizing, just so you can do the integral of sec(x) dx.*0267

*Or else, it is just worth remembering that the integral of sec(x) dx is ln(sec(x) + tan(x)).*0287

*It is a strange and unusual enough integral that is worth memorizing for its own sake.*0296

*Let us go back to our original integral, which I am going to call i, just as we did when we did integration by parts twice.*0302

*What we have now is i has been resolved into sec(x) tan(x) - now we have i again, + that integral that we just did.*0311

*ln(sec(x)) + tan(x).*0328

*Separate that part out and so again just like we did when we had to do integration by parts twice.*0334

*We can move this i over and get 2i = sec(x) tan(x) + ln(sec(x) + tan(x).*0343

*Again, we can solve for the original integral just by dividing both sides by 2.*0364

*We get i = 1/2 of the quantity sec(x) tan(x) + ln(abs(sec(x) + tan(x).*0369

*As always, we have to add the constant at the end.*0386

*So, that is a fairly complicated answer.*0397

*That is a fairly specialized problem.*0401

*When you have a power of sec, an odd power of secant all by itself.*0405

*You want to use this integration by parts, sort of reduction formula,*0407

*To reduce it down to a lower power of secant.*0416

*If you keep reducing it using this formula, you eventually get down to just sec(x) by itself.*0420

*Remember we said that we will memorize that the integral of sec(x) = ln(sec(x) + tan(x).*0429

*That is the end of this lecture on integration of trigonometric functions.*0437

*Hello, this is educator.com and today we are going to talk about integration of trigonometric functions.*0000

*The prototypical examples of these integrals is you will have an integral and some power of sine and some power of cosine.*0006

*The important thing to focus on here is what those powers are.*0015

*In particular, which one is odd.*0022

*This is really a game of odds and evens.*0029

*You want one of those powers to be odd.*0031

*If one of them, if they are both even, then the integral gets more difficult and we will talk about that later in the lecture.*0034

*So right now the important thing is to find one of those numbers and to make it the odd or hope that it is odd.*0041

*If they are both odd, then it really does not matter, you can just pick one.*0047

*The one that is odd, what you do with it, is you are going to make a substitution and let U be the other one.*0051

*U is going to be the other one, and dU will be either + or - the one that is odd.*0076

*It seems a little strange so it would be easier to understand if we work through some examples.*0083

*You will see why this strategy works so well.*0089

*Let us take a look at some examples.*0093

*Here is the first example.*0095

*Cos(x) times sin ^{4}X, dX.*0098

*Of course, cos(x) is the same as cos(x) to the 1.*0105

*So, you look at those numbers 1 and 4, and you see the odd one, that is the odd number.*0109

*What we are going to do is let U be the other one.*0114

*The other one is sin(x) and the point of doing that is dU is then cosin(x) dX.*0120

*So this integral, the cosin, and the dX, become the dU.*0128

*Those two become the dU, sin ^{4}X just becomes U^{4}.*0136

*All of a sudden you have a really easy integral, the integral of U ^{4}.*0144

*Well, that is just U ^{5}/5 + C.*0149

*Then you substitute back, and you get sin ^{5}X + C.*0156

*Just to reiterate there, what made that work was, we looked at which power was odd, and that was the power on cosine.*0167

*Then, we let u be the other one, sin(x)*0177

*The reason we did that is because we had that one power of cosine left over to be the dU.*0182

*So let us try that on a little more complicated example.*0188

*Cos ^{4}X sinU^{3}X dX.*0191

*Again, we look at those two powers and the odd one is 3.*0198

*We are going to let U be the other one.*0201

*U is cos(x) this time, and so dU, well the derivative of cosin is sin(x) dX*0205

*What we can do now is we can take this sinU ^{3} and we can split that up into a sin^{2}X × sin(x) dX.*0207

*The point here is that we save that sin(x) dX to be the dU.*0231

*Well, we are going to have to attach a negative sign, but that sin(x) dX is going to be the dU.*0236

*Then this sin ^{2}, we can convert that into 1 - cosin^{2}x.*0243

*That is why we really had to have this odd even pattern going on.*0249

*It is because we save one sin to be the dU, or actually negative dU, and then you can convert sines into cosines two at a time.*0256

*If we save one sine to be dU, that will leave an even number of sines left over.*0266

*We can covert those all into cosines.*0273

*Our integral, after we make all of those substitutions, cos ^{4} becomes u^{4}.*0276

*The sin ^{2} becomes 1 - cos^{2}, and then becomes 1 - u^{4}.*0285

*Sin(x) dX is dU, or actually negative dU.*0292

*Then we can combine these into one big polynomial.*0297

*Actually, I can bring this negative sign back inside, and make it the integral of u ^{4} × u^{2} - 1 dU.*0304

*Combine those into u ^{6} - u^{4} dU.*0310

*Then, that is an easy integral because then we can just use the power rule, u ^{7}/7 - u^{5}/5.*0321

*Finally, we can convert that back into u as cos(x).*0330

*So, this is cos ^{7}X/7 - cos^{5}X/5, and of course we have to attach a constant there.*0337

*Again, the key thing here was recognizing those powers.*0350

*We see 4, we see 3, 3 is the odd one, so we want to let u be the other one.*0357

*The other one was cos(x).*0365

*The point of that was it gives you one extra sin(x) left over to be the dU.*0367

*Now, we know what to do if either one of the powers is odd.*0373

*What if you have both powers being even?*0382

*That makes it a trickier problem.*0384

*What we are going to use are the half angle formulas.*0386

*If both sine and cosine are even powers, then we can not use the previous strategy.*0391

*That odd strategy does not work.*0398

*We are going to use the half angle formulas.*0401

*Cos ^{2}(x) = 1 + cos(2x)/2*0403

*Sin ^{2}X = 1 - cos(2x)/2.*0407

*The point of those is that those are easy to integrate.*0411

*Even if we have to multiply those together a few times, as we will see in the example, it is not too bad to integrate.*0415

*Let us try out an example of that.*0422

*Here we have sin ^{2}x cos^{2}x dX.*0425

*We look at those powers, we look for the odd one, and oh no, they are both even.*0430

*So, we are going to use those half angle formulas.*0436

*Remember, sin ^{2}x is 1 - cos(2x) over 2.*0438

*Cos ^{2}x is 1 + cos(2x)/2.*0446

*Now I am going to pull the two halves out of the integral to get them away.*0455

*So now that is on the outside.*0460

*Now, we have the integral of 1 - cos(2x) × 1 + cos(2x).*0462

*I will multiply those two together and we will get 1 - cos ^{2}(2x).*0469

*Well, the 1 is going to be easy to integrate, so I am not going to be worried about that.*0477

*But now we have a cos ^{2}(2x).*0481

*How do we integrate that?*0484

*Well, again, that is an even power of cosine.*0486

*We can use the half angle formula again and I am going to write the half angle formula over here.*0491

*But, I am going to use U this time.*0496

*1 + cos(2u)/2 because here we have cos ^{2}(2x), but if you think of that as being U, you can convert that into 1/4 the integral of 1 - cos^{2}(u).*0500

*cos ^{2}(u) is 1 + cos(2u).*0523

*1 + cos, now 2u if u is 2x, u is 4x.*0530

*This is 1/4 × cos(4x) dX.*0540

*That simplifies a little bit to 1/4 × 1/2 - 1/2 sin(4x) dx.*0558

*I am going to pull those halves out and combine them with the 1/4, and we get 1/8.*0573

*Times the integral of 1 - cos(4x) dX, and that is 1/8.*0579

*Well the integral of 1 is just x.*0591

*The integral of cosine is sine.*0596

*But, because it is 4x, we have to multiply on a 1/4.*0601

*Finally, we get (1/8)x - 1/8 × 1/4 = (1/32)sin(4x) and at the end of these we always add a constant.*0606

*In that one, we had to cope with the fact that there were no odd powers.*0626

*Both were even and we used the half angle formula to resolve that.*0632

*Sometimes after you use the half angle formulas, you find yourself with another even power as we did.*0634

*And so you have to use the half angle formula again to resolve that.*0642

*If you keep on using them, eventually you get down to single powers of cosine and that is something that you can integrate without too much trouble.*0647

*Let us look at another example.*0661

*This one is in terms of secants and tangents.*0663

*That is the other common pattern that you are going to see with trigonometric integrals.*0665

*That is, powers of secant(x), and tangent(x).*0671

*To do this, it is helpful to remember that the derivative of tan(x) is sec ^{2}(x).*0674

*The derivative of sec(x) is sec(x)tan(x).*0685

*What we see here is again, looking at these powers, I look at the secant power and I see that that is even.*0699

*The reason that that is significant is that I can separate that out into sec ^{2}(x) tan^{2}(x) dx.*0709

*And, I remember that the derivative of tangent is sec ^{2}(x).*0724

*If I use the substitution u = tan(x) than dU is sec ^{2}(x) dX.*0730

*The point of this is that I can convert this integral into, I have got sec ^{2}(x) dX, so that is dU, tan^{2}(x) is u^{2}.*0744

*Now what about this other sec ^{2}, well remember the other pythagorean identity.*0759

*Tan ^{2}(x) + 1 is sec^{2}(x).*0764

*Sec ^{2}(x) can be written as tan^{2}(x) + 1 which I am going to convert directly into u^{2} + 1.*0773

*This integral converts into u ^{3}, sorry, u^{4} + u^{2} dU.*0786

*Again, that is a very easy integral.*0800

*That just integrates to u ^{5}/5 + u^{3}/3.*0803

*Then we can substitute back, u is tan(x), so that is tan(x ^{5}/5) + tan(x^{3}/3) + C.*0808

*Let us go back and look at what made that work.*0831

*What made that work again was the even power on sec(x).*0833

*If you have an even power on sec(x), then you can always use this substitution u = tan(x) dU = sec ^{2}(x).*0841

*That even power of sec ^{2}(x) guarantees you 2 secants to use as the dU,*0850

*and then the other secants you can convert into tangents using the pythagorean identity.*0856

*The other place you can use substitution like this is if the power on tangent is odd, then you can use the other substitution u equals sec(x).*0861

*Then your dU will be sec(x)tan(x) dX.*0892

*The point of that is that you will have an odd tangent.*0900

*Because you have an odd tangent, you will use this pythagorean identity to convert the tangents into secants two at a time.*0907

*Starting with an odd secant you will be left with exactly 1 tangent left and that tangent you can save that to be your dU.*0918

*We will see another example of that later on.*0930

1 answer

Last reply by: Dr. William Murray

Tue May 13, 2014 4:53 PM

Post by Ying Gao on May 10, 2014

Hi Professor Murray, for the last Additional problem question, I was wondering if when we integrate and distribute - integral (secx (sec^2x -1)dx), why didn't we get -integral sec^3xdx MINUS integral secxdx instead of -integral sec^3xdx PLUS integral secxdx ? I thought the negative sign distributes to both parts?

2 answers

Last reply by: Dr. William Murray

Tue Jan 7, 2014 11:25 AM

Post by Fadel Hanoun on December 19, 2013

For Part one, why can't we say "let u= the even one"?

2 answers

Last reply by: Dr. William Murray

Fri Jun 21, 2013 6:24 PM

Post by Ali Momeni on June 20, 2013

Prof. Murray,

When you took the integral of 1/8 Integral (1-cos 4x) dx did you forget to switch the signs? Should it have been 1/8(x + 1/4 sin (4x) + C ?

Thank you for the excellent lesson otherwise

1 answer

Last reply by: Dr. William Murray

Tue May 28, 2013 7:05 PM

Post by William Dawson on May 22, 2013

I said 'almost' ;p

1 answer

Last reply by: Dr. William Murray

Tue May 28, 2013 7:05 PM

Post by William Dawson on May 22, 2013

God, you make me almost want to ignore the integral tables, lol!

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 11:29 AM

Post by Dr. William Murray on October 22, 2012

Wow, thanks Paula! That's really nice to hear.

Good luck with the calculus!

Will Murray

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 11:30 AM

Post by Paula Hoggard on October 20, 2012

You are an excellent teacher

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 11:32 AM

Post by Riley Argue on June 3, 2012

This video plays correctly.

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 11:31 AM

Post by Riley Argue on June 3, 2012

This video covered everything.