How do I do integration by substitution?

There are a number of integrals that we can evaluate directly, called elementary functions:

integral[x dx] = x2/2 + C

integral[cos(x) dx] = sin(x) + C

However, most integrals cannot be solved so directly. To illustrate the technique of integration by substitution, let’s look at an example.

Example: Find I, where

I = integral[xcos(x2) dx]

It’s not immediately obvious how to start here. But, if you look closely, you can see that there is a part of the function (x) which is the derivative of another part (x2). This is the key to the problem!

The form of the integral should remind you of the chain rule. Integration by substitution is a way of “reversing” the chain rule in a more formal manner.

The fundamental step here is to change the variable of the function in order to simplify it. If we make the substitution u = x2, then we can see that this would make the integral a lot simpler. This is because we know how to integrate cos(u) directly.

However, if we change the variable in this way, we will end up integrating u with respect to x (remember, that’s what the dx bit of the integral means!). Clearly this doesn’t make sense. To get round this problem, we have to integrate with respect to u. Fortunately, we can find this out by differentiating both sides of the substitution:

du/dx = 2x

Now, a word of warning. At this level, we can assume that it is safe to treat du/dx as a fraction, and rearrange it to find dx as the subject. Mathematically speaking, however, this is not the case. So you should be aware that this is simply a method; if you were studying maths at a higher level, you would find a rigorous argument to show that you can perform this calculation.

With that aside, let’s rearrange the expression like so:

dx = du/2x

We are finally ready to put all the substitutions in:

I = integral[xcos(x2) dx] = integral[xcos(udu/2x] = (1/2)integral[cos(udu]

(Note: you can take constant factors outside of the integral sign!)

Hopefully you can now see the point of the substitution, which probably seemed a bit strange to begin with! Let’s solve the integral now:

I = (1/2)integral[cos(udu] = (1/2)sin(u) + C

We’re not quite done yet. Our original integral was in terms of x, but our answer is still in terms of u. We need to change the variable back to x; fortunately, this is easy to do:

I = (1/2)sin(x2) + C


Integration by substitution should be used if the integral looks as though it came from differentiating a function of a function (the chain rule).

We change the variable to u using the appropriate substitution.

We need to remember to find dx in terms of du.

If we are finding a definite integral, we also need to change limits.

Hopefully, this should lead to a solvable integral!

Remember to change the variable back if we are finding an indefinite integral; if we are finding a definite integral, we can just substitute in the changed limits.

Matthew T. GCSE Maths tutor, A Level Maths tutor, A Level Further Mat...

5 months ago

Answered by Matthew, an A Level Maths tutor with MyTutor

Still stuck? Get one-to-one help from a personally interviewed subject specialist


£20 /hr

Monika T.

Degree: Mathematics (Masters) - Exeter University

Subjects offered:Maths


“I'm a Maths student currently studying my first year in the University of Exeter. As a young student who just recently finished A levels and GCSEs, I can remember how stressful they can be and how much work is required for a good grad...”

£20 /hr

Viktoriya L.

Degree: accounting and finance (Bachelors) - LSE University

Subjects offered:Maths, Russian+ 5 more

-Personal Statements-

“Hello, My name is Viktoriya. I am a first year student at the London school of Economics and Poliical science studying accounting and finance. I am passionate about maths and finance in particular and wish to convey this passion to yo...”

£20 /hr

Neal A.

Degree: Mechanical Engineering (Masters) - Manchester University

Subjects offered:Maths


“I'm a student at the University of Manchester and am willing to help every student get a chance to be successful and excel at school and have a bright future.”

MyTutor guarantee

About the author

Matthew T.

Currently unavailable: for new students

Degree: Mathematics (Masters) - Durham University

Subjects offered:Maths, Further Mathematics

Further Mathematics

“Who am I? I am a first-year Maths student at Durham University and have been involved in maths tutoring for a couple of years now.  I’ve worked with students aged between 11 and 18 studying for their SAT’s, GCSE’s and A-levels and  I ...”

You may also like...

Other A Level Maths questions

Discriminants and determining the number of real roots of a quadratic equation

Given that y = 5x^2 - 4/(x^3), x not equal to 0, find dy/dx.

What is the tangent line to the curve y = x^3+4x+5 at the point where x = 2?

Differentiate y^3 + 3y^2 + 5

View A Level Maths tutors

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss