Written by Colin+ in integration.

A few weeks ago, the Mathematical Ninja revealed that he integrated trigonometric functions using a cheap mnemonic. As reader Joshua Zucker pointed out, this was *most* unlike the Mathematical Ninja. Had he been kidnapped? Surely not; no Ninja would ever be taken alive. Had the Mathematical Pirate infiltrated? Had someone just been in a hurry to churn out an article before going on holiday? The truth may never be known.

However, one truth is available: the reasons behind the functions $\sin$ and $\cos$ behaving as they do under integration. There are many reasons. Here are several.

The Mathematical Ninja would, no doubt, call them Euler series. He has, in the past, corrected me on it. It doesn't matter, though.

The Maclaurin series for $\sin$ is that $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... + (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} + ...$

For $\cos$, it's $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... + (-1)^k \frac{x^{2k}}{2k!} + ...$

Where those come from is another post in itself, but the key thing is what happens when you integrate, say, the $\cos$ expression.

You get $\int \cos(x) dx = x - \frac{x^3}{3 \times 2!} + \frac{x^5}{5 \times 4!} - ... + (-1)^{k} \frac {x^{2k+1}}{(2k+1)!} + ... $ - which is precisely the expression for $\sin(x)$1 !

Integrating $\sin$ gives $\int \sin(x) dx = \frac{x^2}{2} - \frac{x^4}{4 \times 3!} + \frac{x^6}{6 \times 5!} - ... + (-1)^{k} \frac {x^{2k+2}}{(2k+2)!} + ...$ - which is $\cos(x)$, although the 1 has been subsumed into a constant. Also, there's a bit of guddling still to do with the general term with all the $k$s in, but it all works out.

If you're lucky enough to be doing Further Pure, you might have come across the functions $\sinh$ and $\cosh$, defined in terms of exponentials. $\sinh(x)$, for instance, is $\frac{e^x - e^{-x}}{2}$; $\cosh$ is the same thing with a plus in between. It turns out that $\sin$ and $\cos$, with a bit of imagination, can be written in the same form:

$\sin(x) \equiv \frac{e^{ix} - e^{-ix}}{2i}$ and $\cos(x) \equiv \frac{e^{ix} + e^{-ix}}{2}$

If you can put aside mental tortures and treat $i$ like the constant it is, you can see that integrating $\sin(x)$ gives you $\frac{\frac{e^{ix}}{i} - \frac{e^{-ix}}{-i}}{2i}$ which, obviously... pardon? Oh.

OK, let's tidy that up: multiply top and bottom by $i$ to get $\frac{e^{ix} + e^{-ix}}{2i^2}$ - which is, of course, $\cos(x)$. You can do something very similar with $\cos$, but I'm fed up of typing the fractions.

The last reason is probably the most obvious: if $\sin(x)$ differentiates to $\cos(x)$, the integral of $\cos(x)$ must be $\sin(x)$. But is it?

Well, of course it is. You can show that using the definition of a derivative and a compound angle formula.

The derivative of $f(x)$ is defined as the limit, as $h$ becomes 0, of $\frac{f(x+h) - f(h)}{h}$. In the case of $\sin(x)$, that gives $\frac{\sin(x+h) - \sin(x)}{h}$, which works out to $\frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$.

Here, it gets tricky. As $h$ gets small, $\cos(h)$ gets as close as you like to 1. Meanwhile, $\sin(h)$ gets closer and closer to... $h$, as the Mathematical Ninja would tell you.

So, the top becomes $\frac{\sin(x) + h \cos(x) - \sin(x)}{h}$. That's good: there's a $\sin(x)$ and a $-\sin(x)$ that obliterate each other to leave $\frac{h \cos(x)}{h}$ - and the $h$s cancel to leave $\cos(x)$. Phew!

The Mathematical Ninja has been warned that his cloak is on a shoogly hook, and that he'd better buck his ideas up in the future.

- give or take an arbitrary constant [↩]

## twentythree

RT @icecolbeveridge: [FCM] Why I Can See My Car works: http://t.co/JntQD4mSLE

## Joshua Zucker

Thanks! This is some great stuff.

## Colin

Thanks!

There was some good stuff with awful proofreading – fixed now (I hope).