The Hardest Integral on Edexcel A-Level Maths?

Added: 2026-05-09

[00:00:00]Okay, this integral is from Edexcel A level math exam in June 2016.

[00:00:05]And this integral does not really look like trig integral, but this radical integral could turn into standard trig integral using simple substitution. That simple substitution is X is 4 * sin squared theta.

[00:00:22]And let's just talk about how and why it is useful to substitute X as 4 * sin squared theta.

[00:00:29]So, look at this integrand. We have X and 4 - X.

[00:00:35]Let's start with this Pythagorean identity. That is sin squared theta plus cosine squared theta is equal to just one.

[00:00:45]And if you multiply 4 to both left and the right-hand side, >> [applause] >> then we should have 4 * sin squared theta plus 4 * cosine squared theta.

[00:00:57]>> [applause] >> This is equal to 4.

[00:01:00]Let me just take this 4 * sin squared theta as the X, right?

[00:01:04]So, choose X as 4 * sin >> [applause] >> squared theta.

[00:01:13]And we can talk about 4 - X.

[00:01:16]Then 4 - X is the same as 4 - 4 [applause] * sin squared theta.

[00:01:23]Okay. Then this is going to be then the same as 4 * cosine squared theta.

[00:01:30]So, then you can talk about this integrand square root of X over 4 - X using this, right?

[00:01:35]Then square root of X over 4 - X is going to be then the same as square root of 4 * sin squared theta over 4 * cosine squared theta. Cancel those 4s out. It is the same as square root of tangent squared theta.

[00:01:55]In in thinking about the limits, right?

[00:01:57]Then the square root of tangent squared theta in our case has to be just tangent theta. So that we can just use it.

[00:02:08]And the hint is if you have square root involving say the form of a squared say a squared minus x, then you can just think about the substitution of x is equal to a squared times sine squared theta. In our case, your a squared has to be just equal to four, right?

[00:02:32]Okay, so using this, let's just work on this integral. I'll be calling this integral as just the I.

[00:02:38]And since we just let x as four times sine squared theta, let's just talk about dx over d theta first, right?

[00:02:45]Okay, then dx over d theta.

[00:02:51]Okay, it is going to be then the same as a times sine theta times cosine theta.

[00:03:02]So that your dx is going to be a times sine theta times cosine theta d theta.

[00:03:15]And since we have square root of x over four minus x as your integrand, that was square root of tangent squared theta, and thinking about the lower bound and the upper bound, it was just a tangent theta. So your integrand has to be then the same as tangent theta times everything.

[00:03:38]Since tangent theta is sine theta over cosine theta, cosine got canceled out, and you have then a times sine squared theta, and then we have d theta.

[00:03:49]So that's why using this, we can just rewrite this integral I as integral. Let's just talk about lower bound and the upper bound, right? But your integrand has to be 8 * sin squared theta d theta.

[00:04:09]About the lower bound and the upper bound, right? So, the lower bound is x is now going to zero.

[00:04:16]And then based on the substitution, x as 4 * sin squared theta.

[00:04:20]But in this case, your theta is also going to zero.

[00:04:23]And the upper bound, when x is going to three, then theta should go to pi over three. So, the lower bound has been now zero, upper bound has to be pi over three.

[00:04:36]And this is your integral now.

[00:04:43]So, the radical integral just turned into standard trig integral. So, we can just evaluate this integral from zero to pi over three, 8 * sin squared theta d theta.

[00:04:52]If you pull this 8 outside of this integral, this is just the same as 8 * integral from zero to pi over three, then of sin squared theta d theta.

[00:05:05]Time to use double angle identity.

[00:05:08]So, sin squared theta is the same as 1 - cosine 2 theta over 2.

[00:05:19]So, if you plug it in, so this is the same as just the 1 over 2 * 1 - cosine 2 theta.

[00:05:27]So, plug it in, pulling this 1 over 2 outside, multiply that with 8, that is equal to 4. So, your integral is the same as 8 * 1 over 2, that is 4, times integral from zero to pi over three, the times 1 - cosine 2 theta, and we have d theta.

[00:05:48]Since we can separate this integral into two integrals, right? Integral of 1 d theta and integral of cosine 2 theta d theta, right? So, if you're talking about that integral of 1 d theta, it is just equal to theta.

[00:06:04]Integral of cosine 2 theta, this is just the 1 over 2 times >> [applause] >> sin 2 theta.

[00:06:13]So, using this is integral of pi.

[00:06:16]Is that the same as 4 times >> [applause] >> bracket of the theta and then minus 1 over 2 >> [applause] >> sin 2 theta and from 0 to pi over 3.

[00:06:33]So, we can just easily evaluate this now, right? It is going to be then the same as 4 times pi over 3 minus square root of 3 over 4.

[00:06:50]Time to distribute this 4 to those two terms inside of the parenthesis. So, the answer for this question is 4 pi over 3 and minus square root of 3.

[00:07:04]This is the answer for the question.

[00:07:08]Okay, so pretty interesting integral from Edexcel A level math exam in June 2016. How amazing.