### Integration Tables

**Main formula:**

Almost every calculus book has tables
of integrals inside the back cover. You can also find tables on the
web. To use these tables, you must make the integral you’re given
match the pattern in the table __exactly__, including the *du*.

**Hints and tips:**

For quadratic expressions, you often have to complete the square and make a substitution to make your integral match the one in the table.

You also often have to make some other substitution like

*u*= 3*x*. When you make these substitutions, make sure you convert the*dx*as well.Sometimes, the tables don’t give you full answers, but instead give you reduction formulas, such as how to reduce ∫ sec

^{n}*x dx*to ∫ sec^{n−2}*x dx*. Then you may have to use the formulas more than once, or use the formula to reduce your integral to another integral that you will then solve directly.

## Integration Tables

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.

## Mathematics: College Calculus: Level II

## Transcription: Integration Tables

*OK, let us try some trickier examples of trigonometric substitution.*0000

*Here we have the integral of 2x - x ^{2} dx.*0006

*The thing that makes that different from the basic forms,*0009

*The basic forms look like a ^{2} - u^{2}, or a^{2} + u^{2}, or u^{2} - a^{2}. *0015

*You will see in your tables of integrals, a lot of those kinds of forms.*0030

*But in all of those, the a is constant.*0036

*The problem with our integral is the 2x.*0038

*It is 2x, and not 2, so that makes it a bit harder.*0042

*However, if you look towards the end of the section of integral tables,*0049

*There is one that covers our situation and I am going to write down the integral formula for you.*0057

*It is number 113 in Stewart's book, it might be a different number in your book.*0060

*But it says that the integral of sqrt(2au - u ^{2}) du = u - a/2 × sqrt(2au - u^{2}) + a^{2}/2, arccos(a - u)/a + C.*0066

*So, we are going to figure out how to invoke that formula.*0100

*The x is acting just like the u in the formula.*0108

*So, we have 2x, and in the formula we have 2au.*0113

*That tells us that the a must be 1 and the u is acting like the x.*0120

*So, we can invoke the formula directly.*0126

*U - a = x-1/2.*0131

*sqrt(2au - u ^{2}) = 2x - x^{2} + now a^{2}/2, a = 1, so that is 1/2 arcos(a-u), is 1-x over a is just 1.*0137

*That can be simplified down to 1-x, plus a constant.*0162

*That one was a complicated formula, but we were actually able to find it directly in the book.*0172

*So that one went a little more quickly than some of the more tricky examples.*0183

*Finally, I would like to mix an exponential function and a trigonometric function.*0000

*The integral we have here is e ^{3x} × cos(5x) dx.*0007

*Again, I am going to use Stewart's book, and he has a section called exponential and logarithmic forms.*0014

*I will show the formula that I am going to be using.*0026

*It is formula number 99.*0030

*So, 99 says the integral of e ^{au} × cos(bu) du = e^{au}/a^{2} + b^{2} × acos(bu) + bsin(bu).*0034

*Again, plus a constant.*0071

*So, let us try to figure out how to reconcile the integral we are given with this formula.*0078

*Well, obviously the a=3, and b=5.*0085

*We can just leave dex/dx as the same as being u.*0092

*Then we can read off the answer from this formula.*0103

*We have e ^{au}, OK, that is e^{3x}.*0105

*Now a ^{2} + b^{2}, that is 3^{2} + 5^{2},*0111

*3 ^{2} is 9 and 5^{2} is 25.*0116

*9 + 25 = 34.*0121

*a = 3 so, 3cosine, b=5, so 5x, + b is 5, 5sin(5x) + C.*0127

*That is our answer there.*0146

*That is the last example on using integration tables.*0159

*I want to emphasize here something that a lot of students do not realize.*0160

*All of these formulas in the integration tables were derived using the techniques we learned in the lectures here. *0170

*So, these formulas should not come as sort of mysterious, magical formulas that you just sort of invoke blindly.*0175

*All of these formulas come from the techniques we have been learning.*0184

*This one that we just used here for example is actually coming from integration by parts.*0190

*We did a problem in the integration by parts lecture where we did integration by parts twice.*0195

*We essentially derived this formula.*0200

*If you are ever stuck without a table of integrals, you can figure out all of these formulas using what you have learned in the other lectures.*0205

*What the role these formulas play is if you do not want to go through the steps for integration by parts twice, *0216

*If you have done that enough times where you understand the procedure,*0220

*And you have to do the integral again,*0225

*You can quickly look it up in an integration table in the back of the book, or on the web.*0230

*You can jump essentially straight to the answer here.*0234

*It is the same answer that you would have gotten had you gotten integration by parts twice, and gone through a little extra arithmetic there.*0237

*This concludes the lecture on integration tables.*0245

*This has been educator.com.*0250

*This is educator.com, and today we are going to talk about how to use integration tables to solve integrals.*0000

*If you are taking a calculus class, you have a calculus book, and if you look inside the back cover of your calculus book, you are going to find several pages of integration tables.*0008

*You can also find these tables on the web. *0026

*We are going to learn to use these tables here today.*0032

*The key thing is that you want the pattern in the integral you are given to match the pattern in the integration table exactly.*0034

*Often that means you are going to have to do a little bit of work with the integral that you are given to get it into a form that matches the integration table. *0045

*We are going to see some examples where you can practice doing that work.*0050

*You get it into a form where you can use the integration table. *0056

*A big part of that is that the integration table will always have a dU.*0063

*You have to make your integral match that, including the dU.*0066

*You will see when we go through the examples.*0074

*So, the first example here, we have the sqrt(x ^{2}+2x+5).*0077

*The calculus book I am using is James Stewart's Essential Calculus, and I am look back at the integration tables there, *0085

*His formula 21 says that if you have the integral of a ^{2}+u^{2} dU. *0094

*Then that converts, or that solves to, (u/2) × sqrt(a ^{2}+u^{2})+a^{2}/2 × ln(u+sqrt(a^{2}+u^{2})+C.*0105

*That is an easy formula to use if we can get our integral to look like that pattern.*0133

*It does not quite look like that pattern yet.*0143

*What we have to do is complete the square on what is underneath the radical sign. *0145

*So we are going to write that as x ^{2}+2x, and then we are going to complete the square on that. *0153

*Remember, completing the square you look at the x term and you divide it by 2 and square it. *0160

*So, 2/2=1 and 1 ^{2}=1, so I am going to write 1.*0164

*But actually, what we had here was x ^{2}+2x+5 so to make that equal I am going to write + 4.*0173

*The point of completing the square was so that we could write this as (x+1)x ^{2}+4.*0185

*We are going to let u=x+1, and then dU=dX.*0190

*You always have to substitute for the dX as well, and so our integral converts into the integral of ux ^{2}+4 dU.*0200

*Now, that matches this formula from the integration table.*0213

*The a ^{2}=4, a=2 and the u=x+1*0220

*Now we can just read off the answer from the integration table.*0230

*Our answer is u/2, that is (x+1)/2 × the sqrt(a ^{2}+u^{2}).*0232

*Well a ^{2}+u^{2} was u^{2}+4 which came from x^{2}+2x+5.*0244

*+a ^{2}/2, that is 4/2 = 2, times ln(u), u is x + 1, + that square root again, a^{2}+u^{2}, came from u^{2}/4 came from x^{2}+2x+5.*0255

*Then we finish it off with a constant.*0280

*The key step there was looking at the integral, identifying which integral formula from the book was going to be useful,*0295

*Or which integral formula from the web if you are looking at the web,*0305

*But then matching the integral we are given to the pattern in the formula in the book.*0309

*The integral we were given does not exactly match, so we had to do this step of completing the square to make it match the formula exactly and then we could invoke the answer.*0315

*So, let us try another example of that. *0327

*We have here arcsin3x dX. Remember, inverse sin is the same as arcsin, not sine ^{-1}.*0330

*If we have a look at the table of integrals, and there is a whole section on inverse trigonometric formulas. *0344

*Formula number 87 in Stewart's book, it might be a different number in your book, is the integral of arcsinu dU = u × arcsindU+sqrt(1-u ^{2})+C.*0361

*Again, our interval does not exactly match the formula from Stewart's book because we have a 3x and Stewart has a u. *0386

*So what we are going to do, is let that u=3x, but in order to make that substitution we also have to convert the dX. *0401

*We have dU=3dX, so dX=1/3dU*0410

*Thus, our integral becomes, I will pull the 1/3 outside, 1/3 × integral of arcsin(3x) and the dX=1/3, we already put the 1/3 outside so now we can say dU.*0420

*Now, it matches the formula from the book perfectly. *0437

*In order to make that work we had to go through the substitution with the 3 and the 1/3, so that 1/3 outside is very important.*0441

*Now that it matches, we can just invoke the formula from the book, so u=3x, arcsin(3x)+sqrt(1-u ^{2}) is 1-9x^{2} + C.*0454

*I should have had parentheses around the whole thing there. *0482

*We can bring the 1/3 in there and get your final answer.*0487

*x × arcsin(3x)+1/3sqrt(1-9x ^{2})+C.*0493

*Again, the key step there is identifying which formula you want to use and then making the integral match that formula exactly.*0515

*Making the integral match that formula exactly, which sometimes involves some substitution along the way. *0521

*Let us try a slightly tricker one.*0534

*This is a trigonometric integral, tan ^{2}x × secx dX.*0536

*I am going to look at the section in the book under trigonometric forms, and I do not see this integral in there exactly. *0539

*But what I do see is an integral, this is formula 71, that says the integral of sec ^{3}u dU = 1/2 secu × tanu + 1/2 ln(secu + tanu) + C.*0552

*So, I do not have exactly the integral of sec ^{3}u, but I know that sec^{2}, I know that I can write that sec^{2}x-1. *0585

*I can write this integral as the integral of sec ^{2}(x-1) × secx dX, which in turn is the integral of sec^{3}x dX - the integral of sec(x) dx).*0603

*Now, I do have sec ^{3}x dX, which is the same as sec^{3}u dU, except that you are just substituting x for u. *0629

*I can write this as 1/2 secx × tanx + 1/2 ln(secx + tanx).*0640

*I will add the constant at the end of this when I am all done.*0657

*In the mean time, I still have to cope with this integral of sec(x), *0661

*But we learned in the lecture on trigonometric integrals that we can solve that one by multiplying top and bottom by tan(x) × sec(x),*0664

*The point of that was that it solved the whole integral into the ln(secx+tanx). *0674

*This actually simplifies a little bit, 1/2 secx × tanx, now we have 1/2 ln of all that minus ln of all of that.*0699

*We can combine those into 1-ln(secx+tanx).*0712

*Now I will add on the constant because we are done with the whole integral.*0721

*Just to reiterate here, you might not be able to invoke the formulas from integral tables immediately.*0733

*Sometimes you have to take the integral you are given and do a little bit of manipulation on it the way we did with the tan ^{2}, *0740

*And convert it into some form where you can use an integral table.*0745

*You do that, you may get extra terms, like we had the extra term of the integral(secx), *0752

*You still had to remember how to integrate that from what we learned in the trigonometric integral lecture.*0758

1 answer

Last reply by: Dr. William Murray

Mon Mar 9, 2015 9:46 PM

Post by Amanda so on March 7 at 04:17:32 PM

Hi Professor, Im stuck on my homework integration problem, wondering if you can help me find out what I'm doing wrong:

The problem is:

integral sec^3(5piex) dx

I chose entry 71:1/2 sec(u)tan(u)+1/2ln|secu+tanu|+C

The result I get is

1/2pie sec(5piex)tan(piex)+1/2pieln|sec(5piex)+tan(5piex)+C.

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 12:28 PM

Post by Shehan Gunasekara on May 16, 2012

Couldnt see what the integration tables look like???!!!!!!!!!!!!!!!!!!!!!!!