Until fairly recently, I had always done the kind of differential equations you see in Core 4 the same way: separate, integrate, substitute, celebrate. I have taught any number of students the dance; many of them have boogie-woogied their way into a correct answer in exams.

But there’s a variation I’ve started liking more – in the penultimate step, rather than substituting in the values you’re given to work out a constant, instead use them as the lower limit of integration; the upper limit is the $x$, $y$, $t$, or whatever variables you’re working with.

Confused? Good. Let’s try one:

$\diff{y}{x} = 2 \cosec(2y) \sec(y) e^{x}$, and when $y = \frac 14 \pi$, $x = 0$.

The separation step works as always: multiply both sides by $\sin(2y)$ to deal with the cosecant, by $\cos(y)$ to deal with the secant, and by $\dx$ ((Hush, purists.)) to leave you with $\int \sin(2y) \cos(y) \d y = \int e^{x} \dx$.

It’s at this point that you can put your limits in:

$\int_{\frac{\pi}{4}}^{y} \sin(2Y)\cos(Y) \d{Y} = 2 \int_{0}^{x} e^{X} \d{X}$ ((The letters in the integrands ought to be capitals – or at least different letters from $x$ and $y$ – for technical reasons; if you don’t bother changing them, I won’t tell anyone.))

The left hand side works out nicely: $ \int_{\frac{\pi}{4}}^{y} \sin(2Y)\cos(Y) \d Y = \int_{\frac{\pi}{4}}^{y} 2 \sin(Y) \cos^2(Y) \d{Y} = \left[ - \frac {2}{3} \cos^3(Y) \right]_{\frac{\pi}{4}}^{y} = -\frac{2}{3} \left[ \cos^3(y) - \frac{1}{2\sqrt{2}}\right]$.

The right hand side is simpler: $\int_0^x 2e^{X} \d {X} = \left[ 2e^X \right]_0^x = 2e^x - 2$.

Matching them up gives: $\frac{2}{3} \left[ \frac{1}{2\sqrt{2}} - \cos^3(y) \right] = 2 e^x - 2$.

A quick tidy-up and rearrange gives: $ x = \ln \left( \frac{1}{6\sqrt{2}} - \frac 13 \cos^3(y) + 1 \right)$.

The traditional way would have had:

$-\frac{2}{3} \cos^3(y) = 2e^x + C$

Substituting in $y=\frac {\pi}{4}$ and $x=0$:

$-\frac{2}{3} \frac{1}{2\sqrt{2}} = 2 + C$

$C = - \frac{1}{3\sqrt{2}} - 2$

So $-\frac{2}{3} \cos^3(y) = 2e^x - \frac {1}{3\sqrt{2}} - 2$

And still a fair amount of rearrangement to do before getting the same answer as before.

Is the ‘limits’ method better than the ‘constant’ method? I think that depends on what you’re happiest with. Personally, I find that I make fewer mistakes with the limits method, and that there’s a certain elegance to having your constant show up as a result of the integration rather than through explicitly putting it there.

However, I’m conscious that there’s a rabbit-hole here: if it’s good enough for differential equations, why isn’t it good enough for integration in general? That’s a very good question, and part of me wonders if that’s exactly how we should teach integration in general.

For example, the differential equation $\diff yx = m$, given that $y=y_0$ when $x=x_0$ becomes:

$\int_{y_0}^{y} \d Y = \int_{x_0}^{x} m \d X$

$ \left [ Y \right]_{y_0}^{y} = m\left[ X \right]_{x_0}^{x}$

$y - y_0 = m(x - x_0)$, as God intended.