This is just a quick reminder. If you find any of this confusing, there is a very important trick for making it easier – practice, practice, practice! It doesn’t take years to master this, it takes a few hours every week for a few weeks. You will become more and more familiar with the techniques and learn intuitively to know which technique to use in which situation.
Let’s start with the very basics.

If F(x)=x^2, what is its derivative? ie. how do we find a function f(x) such that:

f(x)=\frac{d (F(x))}{dx}


The answer of course is f(x)=2x. We use our normal differentiation rules which you should now be very familiar with. How about if I told you that there was some function F(x) whose derivative was 2x – ie. we reversed the question:

What is a function whose gradient at a point x is 2x?

How do you find what F(x) is? We are trying to solve the equation

2x=\frac{d F(x)}{dx}


for F(x). The answer to this is F(x)=x^2+c. It’s easy to plug it back in and see that indeed this is correct – take the derivative and see that it does give 2x – when we differentiate the constant, c, we get zero. We can say that the antiderivative (or indefinite integral) of 2x is x^2+c and write:

\int 2x dx=x^2+c


If the derivative of F(x) is f(x) then the indefinite integral of f(x) is F(x)+c. Actually here it’s easy to see where the c comes from. The original question was: What is a function whose gradient at a point x is 2x? Well, we can see that if we find some function which satisfies this, then adding a constant to that function which raises or lowers the whole function by that constant, then it won’t change the gradient.

It will always be easier to check whether the answer to an integral question is correct than finding the answer in the first place.

If you think you have correctly calculated an integral – check it!! Differentiate it and you should get the INTEGRAND (that function being integrated).

You should be extremely familiar with the basic rules of integration by now. If not, practice over and over again until they are natural.

Review:Integration by substitution

The trick is to become familiar enough with the examples to know what substitution will make the problem solvable. There are a fair number of tricks to know, which can’t all be taught in class, but you can get them by studying the problem sheets in detail. It takes some practice but you will, over time, understand WHY what works works.

What is:

\int \cos(x^2) 2x dx


What if we substitute u=x^2:

\frac{du}{dx}=2x\,\, \Rightarrow\,\, 2x dx=du


Let’s substitute this into the original integral:

\int \cos(u) du=\sin(u)+c


we then substitute back again for x to get:



We can note something important about the original integral:

\int \cos(x^2)2x dx


2x is the derivative of x^2 which appears in the function \cos x.

Warning: This is going to get a bit abstract, but work through it line by line and you will understand.

Let’s say that f(x)=\cos(x) and g(x)=x^2. Then we can write what we have as:

\int \cos(x^2)2x dx=\int f(g(x))g'(x)dx


Check that you agree with the above

But we know from the chain rule that:

\frac{d F(g(x))}{dx}=F'(g(x))g'(x)


Again – make sure that this makes sense.

We can integrate this equation to get:

\int \frac{d F(g(x))}{dx} dx=\int F'(g(x))g'(x) dx


we are just integrating both sides of the above equation – nothing fishy going on here!

but \int \frac{d F(g(x))}{dx} dx=F(g(x))+c


The integral of a differential is the differentiated function, plus a constant

which means that:

\int F'(g(x))g'(x)dx=F(g(x))+c


or alternatively:

\int f(g(x))g'(x)dx=F(g(x))+c


We’ve just renamed f(x)=F'(x) ie. F(x) is the integral (antiderivative) of f(x).

Now let’s go from this abstract example to the concrete example with f(x)=cos(x) and g(x)=x^2.

The antiderivative of \cos(x) is \sin(x)+c and so in the above example F(x)=\sin(x)+c. Now we have:

\int \cos(x^2)2xdx=\sin(x^2)+c


because the answer will be F(g(x)) as we saw above.

If this doesn’t make sense, run through it a few times and see why we’re making these substitutions.

OK, so we just saw a particular example of an indefinite integration by substitution. Here are some more for you to try…

How about a definite integral by substitution:

\int_{e^1}^{e^2}\frac{1}{x(\ln x)^3}dx


This is really shorthand for:

\int_{x=e^1}^{x=e^2}\frac{1}{x(\ln x)^3}dx


Try the substitution u=ln x (ie. F'(x)=\frac{1}{x^3}, g(x)=\ln x in our explication above.):



which means that:



We have to be careful that now we will be integrating over u and not x, so we have to change the limits too. When x=e^1, u=1 and when x=e^2, u=2. So the new integral is:

\int_1^2 \frac{1}{u^3}du=\left.\frac{-1}{2u^2}\right|_1^2=\frac{3}{8}

Looking of the graphs written in both the x variable and the u variable they look very different. The point is that the area is still the same. Both the integrand AND the limits change to get the same answer.

Go here for many examples – work through them and you will master this technique!

How clear is this post?