Is there a simple way to integrate an expression such as

x² (x³ + 2)⁷ dx

without using integration by parts 7 times or expanding using binomial theorem then integrating term by term?

Context: this is supposed to be of the level of London Pure Maths P2 and is taken from Delphis Publications A level section tests.

Bill Bradley

There is a method of integration which you can use here which is often known as "integration by recognition". The idea is that you try to recognize what you would have to differentiate to get what you want to integrate.

Differentiating (x³ +2)⁸ by using the chain rule gives 8(x³ +2)⁷ *3x² = 24x² (x³ +2)⁷ , which is nearly the thing you want to integrate except for a factor of 24.

Therefore, the answer you require is (1/24)(x³ +2)⁸

Basically, this method relies on spotting that the term outside the bracket is roughly the derivative of the thing inside the bracket, apart from a factor which you must divide by.

Jo

Dear Bill,

There are 3 main ways to integrate at A-Level, integration by Parts, integration by Substitution, and what my friend calls integration by `Spot The Ball'! All three methods will work here, but, as you've found out, parts is an absolute nightmare. I'll explain in more conventional terms what `Spot The Ball' is, then deal with the integration.

You'll know that if you've got a bracket raised to a power, such as (2x² -10)^7, and you want to differentiate it then, rather than expanding and differentiating term by term we can do it by substitution: I.e., let u equal what's in the bracket, u=2x² -10, in this example. Then use the formula d/dx = du/dx d/du. (In words, as it's not very clear,: d by dx equals du by dx times d by du.) If you write the right hand side down, you'll see that you can almost `cancel' du from top and bottom to leave d/dx on both sides. Although this isn't strictly true, it's a VERY helpful way to think of it. So now, instead of differentiating e.g. (2x² -10) with respect to x, we have to differentiate u⁷ (this is our substitution) with respect to u, then times by du/dx.

d((2x² -10)⁷ )/dx = du/dx d(u⁷ )/du

These are a lot more simple to work out:

du/dx = 4x, d(u⁷ )/du = 7u⁶ .

Ie: d(2x² -10)⁷ / dx = 28x(2x² -10)⁶ .

(Notice we get an x outside the bracket) Now integration by substitution is very similar, and Spot The Ball is really just a fast way of integrating by substitution once you're a bit more experienced.

If we want to integrate x² (x³ +2)⁷ , we think about it a bit, then miraculously come up with the substitution u=(x³ +2). (It gets easier to spot which substitution to use when you've had more ractice). However, we also need to deal with the dx which is in the integral, and we do it by using:
dx = dx/du x du (Notice like above that the du's `cancel'?)


Now du/dx = 3x² and using the fact that dx/du = 1/(du/dx) (another one to remember!) we can see that the integral is transformed: x² (x³ +2)⁷ dx = u⁷ /3 du. But that's easy to calculate: u⁷ /3 du = u⁸ /24 + constant.

= (x³ +2)⁸ /24 + constant


Substituting back in at the last step.

Can you see how we could have spotted this at the beginning? Notice x² (x³ +2)⁷ is the derivative of (x³ +2)⁸ /24. And since integration and differentiation are opposite processes, the integral of a derivative is the the thing we have differentiated. So this would have given us the same answer as before. This requires that we can `spot' when we are integrating something that has been differentiated, hence the apt title `Integration By Spot The Ball' (or just boring reverse differentiation to your teacher).

This method is very useful for integrating things without having to use parts, which can take AGES! There are loads of examples of this, and it works with lots of other functions as well. For example, $2x{e}^{x^2}$ (look at the derivative of $e^{x^2}$, and us a substitution if necessary), or $\cos x(\sin x{)}^5$.

Hope all of this helps, I've gone through it all at quite a length, but hopefully you see how integration and differentiation are so closely linked. If you want to look it up in a text book, this type of differentiation is often called differentiating a `function of a function', or `chain rule', and the integration is integration by substitution. (I doubt you'll find a book where it's called Spot The Ball, but I thought it sounded really good!) Finally, I hope your exams go well. I must admit I used those Delphis papers when I was revising for my exams, and I thought they were really hard, so rest assured you're not the only one!

Best Wishes,

James Walsh.