 ###### John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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# Integration: Substitution Method

John Postovit ###### John Postovit

University of North Dakota
M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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[0:00:00]
There are a number of techniques for dealing with more complicated integrals from integration by parts, to trigonometric substitution, to the substitution method. That last one, that's one of the most important. You almost certainly have to use it. It's considered relatively basic. What the substitution method does is, it's really cool I love this method. It basically undoes the results of doing a derivative with a chain rule. And I usually just call it UDU. It kind of got a nice sing song touch to it that way.

When the substitution method works for a problem, it's just beautiful. It works like a charm. Just fantastic. Let's take a look at the steps. In the substitution method the first thing you do is you pick a likely value of something, a likely section of the integral that you're going to define, is u. That's where the art of this comes in. You're going to decide what u is and sometimes that can be a little bit tricky until you've practised. Well, then you take it's derivative and if you take it's derivative you have to solve that derivative for du.

The reason you do that is because you're going to need to substitute. Now if you chose u correctly, all the parts of u and du, all the variable parts are going to show up in the integral and be ready to do the substitution. If not, you have to pick another value for u and try again. Finally, after you've substituted, you integrate it and then you re-substitute the original variables. Now let's do some practice problems involving the substitution method.

Remember our first step was to choose a likely value for u. That's where again the art comes in. You might say, "I want u to be 3xÂ²." I'm going to pick that because that's not it, I'm going to show you what happens if you pick the wrong thing.

[0:02:00]
If you decide that you want u to be 3xÂ² and then you go and take u=3xÂ² and do d/dx to both sides. So this side you'd have du/dx equals, now d/dx of 3xÂ² is 6x. And then you can actually treat that dx like it's a denominator. Multiply both sides by dx, and you get your du substitution. It is 6xdx. After you have done this a few times, you don't normally write out all these steps. You usually just go to this one, because you know you're going to be solve for dx anyhow.

Now here's the problem, we've got our 3xÂ² we could substitute that 3xÂ² with a u, but we don't have anything for the du. You have to have another 6x in there to make it happen, and it's not there. And it's got all this other stuff too. So we picked the wrong value for that. Let's try again.

See this 7th power, often times when you have something really complicated underneath a power or underneath a root, that's going to be your choice for u. So I'm going to pick u to be xÂ³+1, not the 7th power just the xÂ³+1. And if I do du on that, then I wind up with du equals, derivative of xÂ³+1 is 3xÂ². And then there is that dx factor that comes from going through all these steps. Now we can go to the tell and look. This is 3xÂ² and the dx.

[0:04:00]
Together those two make up du. And underneath that 7th power, is the xÂ³+1. This time I'm going to rewrite the order on it, because often times the combination of different parts to make du can confuse you. So rewriting the integral, I'm going to put the xÂ³+1 to the 7th power first. And then I'm going to put the 3xÂ²dx. So that's times 3xÂ² dx. So notice, I just took the integral and changed the order of multiplying, you can always change the order of multiplying.

This section right here is entirely identical to what du equals. So I can replace it with du. This section right here does have something that u doesn't, but that's okay. I can replace this stuff in the parentheses with u, and then the 7th power will be on u, not on me. That was a joke.

Now look at that, nice, easy integral. Remember for basic integration, you're undoing a derivative. When you undo a derivative, you have one more power. Remember a derivative takes the power in the front, off the front as a factor and then reduces the power by one. So one more power then u to the 7th is u to the 8th, and also remember, there is always a correction factor with integrals. Because if you were to do the derivative u to the 8th, you would have 8u to the 7th. And there is no 8 in there.

The only way you cannot have that 8 is if we multiply it by 1/8.

[0:06:00]
If I did that derivative, that 8 would multiply the 1/8 getting rid of the factor and then we just have u to the 7th. Now that little detail that you don't want to ever forget, especially on the free response section, because they will dock you for it, plus constant. Whenever you undo a derivative, you have a plus constant. Unless you're doing a definite derivative where your definite integral or you're substituting in the numbers. See if I did the derivative of this, the derivative of any constant is 0, no matter what it was. If the constant was 6 million, the derivative of that will still be 0, and it wouldn't have shown up in what's underneath the integral.

All that being said, we're almost done. All I need to do is the last step in the substitution method, which is to re-substitute for u. So our final result is 1/8, u was xÂ³+1, plus your old friend the constant.

In this next problem, again your difficulty is deciding what u is going to be. Now you could make u the cosine of 3x, because the derivative of cosine 3x is sine x. The problem is, that sine is in the denominator. So if you chose cosine 3x, you wouldn't have the parts you need. What I'm going to do is choose sine 3x as my value for u. u equals sine of 3x.

Now when I do du on this, the derivative of the sine is cosine, don't forget though this is a problem where you're going to do the chain rule. So the derivative of 3x is 3and du works out to be 3 cosine 3xdx.

[0:08:00]
Now let's see what we have. I'm going to re write this one again, to make it more obvious what's going on. I can take that fraction and split it up into a fraction times another part. We would have 1 over square root of sine 3x times the cosine of 3xdx. Cosine of 3xdx equals du.

Now one temptation you have to avoid is, say you do du and it turns out that some of the components you need are missing. Sometimes that's okay, sometimes it's not. We don't have the 3 that we need, that is okay, because it's a constant. But if du ever requires you to have something that's a variable, that's not there, you can't put it in. You can't add variables inside an integrals. You can put in constant factors and that's it. So I do need a 3 so that I can replace with du. I'm going to multiply by three here. I guess I have to put it out front. Of course you can't multiply something by 3 without making it larger, so I have to undo that by also multiplying by 1/3. So we put in an extra factor that we needed and then we correct it for doing that.

This 3, and the cosine 3xdx all together those factors equal to du, replace it.

[0:10:00]
And 1 over the square root of sine 3x. Sine 3x is u so I can do my substitution there, that's 1 over square root of u and remember the 3 is gone, but there is still the 1/3 out front. Now we're ready to integrate. Before I integrate, when you have things in the denominator, it's really good to make it be fractional powers, make it be polynomials. 1 over the square root of u is the same thing as 1 over u to the 1/2, which is the same thing as u to the -1/2.

So this is 1/3 of the integral of u to the -1/2 du. The reason I did that is because integrals are easier to do, if you can just count up on the powers. And even this can be a little tricky. But just remember, you're adding one to the power, if you take -1/2 and add 1 to it, you have +1/2. This is the 1/3 factor that we needed for our correction. One more power then u to the -1/2, is u to the +1/2. All right then don't forget the basic correction factor for integrals.

If I did he derivative of u to 1/2, that will be 1/2 of u to the -1/2. But there was no 1/2 in there, so I need to correct for that by multiplying by 2, that's the correction factor, plus your friend the constant. And the last thing to do, is to re-substitute for u. Our final result on this one, 1/3 times 2 is 2/3, times sine 3x to the 1/2 power.

[0:12:00]
I'm going to make the 1/2 power to be a square root because we started with square root, radical form in the regional problem. It just needs a notation. We have the square root of the sine of 3x plus your old friend the constant.

Now have a look at this one, this is sneaky. You see all the stuff underneath the square root and you think, "Men we're in luck, we can use substitution, we can use UDU." Yeah. But if you did, it's not going to work. If you decide to make u be 2xÂ³-7, then du works out to be 6xdx. Then there is no 6x in there and you can't put it in, because it involves the variable. This is pretty sneaky. It's a binomial just square it.

Remember when you square a binomial, you square the first thing. (2xÂ³)Â² is 4x to the 6th power plus the first term times the second term doubled, so 2xÂ³ times -7 times another 2 is going to give us -28xÂ³. And if you're squaring -7 you get +49. Look at that beautiful it's just a plain old polynomial to integrate.

Remember when you have a polynomial that you're integrating, you can actually integrate each subsections separately, which is effectively what we're going to be doing because we're going to need different correction factors.

[0:14:00]
So for this time I'm actually going to write this one out. We have 4x to the 6th minus the next integral, which
would be -28xÂ³. Whenever you have a constant factor, remember that you can pull it out the front as a multiplier, a regular factor. And yeah I'm applying dx to all of then because we're integrating all of them. Each one is a separate integral now. The last one is the integral of 49dx, I could have pulled the 49 off front but I didn't because I find it more confusing that way to have nothing behind the integral except for the dx. We're close to done. One more power then x to the 6th is x to the 7th, and that correction factor for that is the correction for the derivative. Derivative of x to the 7th is 7x to the 6th. So I need a correction factor of 1/7. Don't forget that there is still this 4 in there that I didn't pull out in advance, even though I could have.

Next part would be -28 times the integral of xÂ³ is power more, that's x to the 4th. And the correction factor for the 4th power is 1/4. The last one, integral of 49. 49 is a constant, that's 49 times x to the 0 power, one more power that x to the 0 is x to the first. 49 x to the first, This time we don't need a correction factor because the derivative of x to the 1st is just 1, plus you're old friend the constant.

[0:16:00]
You can count on having to do the substitution method on the AP test, pretty much guaranteed. The thing I'm going to show you next, change of variables is a little bit more advanced. It's one more step on top of the substitution method. You might not have to do this one, but it's best to be prepared.

So it's got a little bit more work to it. You start by taking a likely section of the integral to define as u, take the derivative but then you solve the u for x. And you solve du for dx. Then you look to see if you're on the substitutions you need there. You substitute or if that doesn't work out you choose another u if needed, then you integrate it and re-substitute it. Okay this problem here is one where you're looking at this and deciding, "I think I probably need the substitution method." And then you go and choose u=3x+1. But after you do that then du will be just 3dx and you wouldn't have that 2x to the 4th factor that you need. So straight substitution method is not going to work. But change of variables, that will. I'm still going to choose u=3x+1.

Typically again anything underneath some other functions tends to be u. So I'm going to pick u=3x+1, and then I'll find out what du is. This time I'm going back and going all the way to the basic method that I didn't use after the first time I showed you. I'm doing du/dx to both sides. So du/dx to both sides is just 3, because the derivative of 3x+1 is just 3.

Next, this is where it differs from basic substitution. If you substitute now you won't have the parts that you need and what I need to do is solve both of those for the x.

[0:18:00]
So if I take u=3x+1 and solve that for x, you'd take 1 from both sides. You divide both sides by 3. And you would have x=(u-1) over 3. With du/dx if I cross multiply that, I'd have du equals 3dx but then I solve that for dx. And I'd get dx=1/3 of du, that's not a negative sign there, that's an equal. So dx is 1/3du.

Now this doesn't look very promising, because it looks so complicated. We're winding up with these fractions. But watch what happens when I do the substitution. So let's see, that 2 factor can be pulled up front. It is always a good idea to do that because it's going to help you avoid getting factors like that entangled with correction factors.

Next, I've got to replace that x with (u-1) over 3. This is a tight face, it's a little bit hard to read here. That's a 4th root, not 4th power on the x. So the x gets replaced with (u-1) over 3 times, I've got the 4th root of 3x+1. I'm going to write that though as a 1/4th power. Now everything underneath there, 3x+1=u, we made it be that way.

[0:20:00]
So that equals u the radical mix of 1/4 power. This entire unit right here is u to the 1/4 power. Now remember the substitution method, when you're getting rid of the dx, you did that by finding a combination of factors that made up du. But this time, all I have left, the only thing I haven't substituted for is dx. And I'm going to replace the dx with 1/3 of du.

Now at this point, we have to clean this thing up. I have got 3 is in the denominator. They're just constant factors, so I can pull them out. 3 times 3 is 9. So I wind up with a 2/9 factor in front. And then I have this u-1 times u to the 1/4 power. This is starting to look nice. We don't now need to do a substitution again. You can distribute that 1/4 power into the parentheses. And now we have 2/9 times the integral, u to the 1/4 times u is u to the 5/4, you know 1 + 1/4, u to the 5/4. u to the 1/4 times -1 is -u to the 1/4. We've got du on it. Polynomial easy to integrate.

Now unlike the one I did a minute ago, I'm not going to take the time to separate it out into different integrals. I'm just going to deal with it as a group of parts. So I do need to remember that when I do this, that the 2/9 has to multiply everything. Integral of u to the 5/4. 5/4 plus 1 is 9/4.

[0:22:00]
So I have u to the 9/4th. If you do the derivative of u to the 9/4th, that's 9/4, u to the 5/4 but there is no 9/4 in there. Ignore that outside because 2/9 has nothing to do with. We still need the correction factor of 4/9 minus the integral of u to the 1/4 is one more power than that, which is u to the 5/4. The correction factor for that is 4/5, plus your old friend the constant.

I know I put the constant on the outside bad me, it has actually turned out to matter. Technically it should be inside because those brackets included the integral, but in the long run it doesn't matter. It doesn't matter because if you multiply a constant by 2/9, you get just another constant anyhow. You don't care what the constant is, it's just a number.

Before we can lay this problem to rest, there is one last thing to do. You have to replace the u with of course 3x+1. Don't forget to do that. I won't do it in this case. but if you look ahead into the bonus materials, you'll see it all written out for you.

Bear in mind that even though you use the substitution method all the time, you'll never going to find a problem that says, "Do the substitution method for this problem." You're going to find it buried under problems. For example if you take a look at the practise problems for differential equations, you'll see one of those requires you to do the substitution method, in order to solve the differential equation.

Well, before we go I've got something kind of like fun for you. Caltech gets the credit for it, it's known as the caltech calculus chart. No one ever knows where it really came from but it's great.

[0:24:00]
Can you just imagine marching in to the AP test, lock-step singing the calculus chart. So let's go caltech calculus chair. E to the x, e to the x, e to the x, e to the x, 3.14159, secant cosine tangent sine, calculus calculus raaah!