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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???!!!!!!!!!!!!!!!!!!!!!!!

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 = 3x. 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 ∫ secn x dx to ∫ secn−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 - x2 dx.0006

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

The basic forms look like a2 - u2, or a2 + u2, or u2 - a2. 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 - u2) du = u - a/2 × sqrt(2au - u2) + a2/2, 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 - u2) = 2x - x2 + now a2/2, 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 e3x × 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 eau × cos(bu) du = eau/a2 + b2 × 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 eau, OK, that is e3x.0105

Now a2 + b2, that is 32 + 52,0111

32 is 9 and 52 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(x2+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 a2+u2 dU. 0094

Then that converts, or that solves to, (u/2) × sqrt(a2+u2)+a2/2 × ln(u+sqrt(a2+u2)+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 x2+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 12=1, so I am going to write 1.0164

But actually, what we had here was x2+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)x2+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 ux2+4 dU.0200

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

The a2=4, a=2 and the u=x+10220

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(a2+u2).0232

Well a2+u2 was u2+4 which came from x2+2x+5.0244

+a2/2, that is 4/2 = 2, times ln(u), u is x + 1, + that square root again, a2+u2, came from u2/4 came from x2+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-u2)+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=1/3dU0410

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-u2) is 1-9x2 + 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(1-9x2)+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, tan2x × 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 sec3u dU = 1/2 secu × tanu + 1/2 ln(secu + tanu) + C.0552

So, I do not have exactly the integral of sec3u, but I know that sec2, I know that I can write that sec2x-1. 0585

I can write this integral as the integral of sec2(x-1) × secx dX, which in turn is the integral of sec3x dX - the integral of sec(x) dx).0603

Now, I do have sec3x dX, which is the same as sec3u 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 tan2, 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