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7d. Integration by Substitution

Integration Mini Video Lecture

This video explains how to use substitution to make an integral that looks difficult, into one that is easier.

Video transcript


Related lessons on IntMath

You will use the techniques in this video later, in the following lessons:

Integration: The General Power Formula

The Basic Logarithmic Form

The Exponential Form

Integration by Trigonometric Substitution

Let's start by looking at an example with fractional exponents, just a nice, simple one.

Before I start that, we're going to have quite a lot of this sort of thing going on, where we get some kind of fraction on the bottom of a fraction, and it gets confusing.

So what we should do is think about this as, like, 1, divide, and then 3 over 2.

When I divide by a fraction, what I should do is multiply by the reciprocal, and this just gives me two-thirds `(2/3)`.

Now, here's a simple integration here, involving a fraction. So when I'm integrating this sort of thing I should add 1 to the index and then divide by the new number.

So I'm going to have something like this: I'm going to have on the top v and then

... 3 over 2 `(3/2)`,

... over 3/2 - because

... a half `(1/2)`

... plus 1 is "1 and a 1/2" `(1 1/2)`,

... and then divide by the new number - plus K.

Now, this thing looks a bit untidy, but I can use what I had up here, I can just type this as 2/3 of v to the 3/2 `(2/3 v^(3/2))`, and then, plus K.

Now, normally I don't even do it like this. I don't have this step in the middle, I just go straight to this, yep, add 1 to the index here, you get 3/2, turn it upside down and you hit 2/3 times `v^(3/2)` plus K.

You put this step in the middle if you're not sure what's going on.

Great. Let's move on to the main problem now.

What we want to do is we want to integrate the square root of 5x minus 4 `(sqrt(5x-4))`.

So the first thing we should do here is to realize that `sqrt(5x-4)` can be written as 5x − 4 in brackets, to the power 1/2, and then put dx.

Now, with this sort of problem, we should do some substituting. And the normal substitution to do is u equals 5x minus 4 `(u = 5x-4)`.

I've chosen 5x − 4 because that's what's in the brackets here.

The step that you do after that is always `(du)/dx`, and `(du)/dx` in this case is 5.

Now, as part of the substitution, I'm going to replace this with u, and this will become u to 1/2 `(u^(1/2))`, but I've got a dx here, and I can't have an extra u and dx, I've got to do something about it.

So I write the expression I've just done, `du = 5dx`, and one more step I'm going to have one-fifth [1/5] of du equals dx `(1/5 du = dx)`, and all I've done now is divide both sides of this by 5.

Okay. Now, what I've got to do next is do the substitution steps. So I take this and just paste it in here, and now really put the substitution: 5x − 4 (I've said it's going to become u, and 5x − 4 was to the power `1/2`, so I must do it to the power `1/2`.

dx is equal to `1/5 du`, so I do a du, and where should I put this `1/5`? Well, the best place to put it is out in the front, like this. It's just a constant, and so I can just put it out in the front.

So here I've got some expression involving x with a dx. This is correct. I've got an expression with u and du. This is also correct. In this work, don't mix up your x's and du's - it gets very confusing and probably incorrect.

Okay. So I just do that `1/5`, the end, and now what I have to do is I have to integrate `u^(1/2) du`.

But this is just like the problem I did above, isn't it? Let's go and have another look quickly. Yes, look.

My integral of `v^(1/2) dv` and I did it like this. Well, do the same thing down here, integral of `u^(1/2) du` is going to be `2/3` what we got before, `2/3 u^(3/2)` and then plus K. The `3/2` comes from adding 1 to this, 1 plus 1/2 is `1 1/2` or `3/2`, and the `2/3` comes from dividing 1 by `3/2` or multiplying by `2/3`.

Then, let's tidy this up, we're going to have fraction out in the front, `2/15`, and it's going to be `u^(3/2)` plus K. Now, we're almost finished, but not quite. What we want to do now is express this back in terms of x. If we don't do that, that's incorrect - we started with x, we should finish with x. Now, it's always a very good idea to have a clear statement of u, because you always have to substitute back again.

I go and look up here, here is my clear statement of what u is, and so I substitute back into here: 5x − 4 and raise it to the power `3/2`, because that's what it is, and plus K. And that's the end of the story.


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