Integrating Trig Functions

Okay. I HATE INTEGRATING TRIG FUNCTIONS. Absolutely hate doing it. But I have to for Spec. So I'm going to explain how to do it.

The Basics

Probably one of the main reasons why I hate integrating trig functions is because I keep mixing up the derivatives of the trig functions with their integrals. Plus you might get some that seem tricky to work out and then you realise that the trig function in question is simply a multiple of one of the three derivatives of tan x.

Here are the integrals that you need to know:

Integral of sin x = -cos x
Integral of cos x = sin x
Integral of (1/(cos x)^2), (sec x)^2 or 1 + (tan x)^2 = tan x

All the other rules work as before. If the trig function's raised to a power, raise the power by one and divide by the new power and the derivative of the trig function- but make sure that a multiple of said derivative is also present. So you can't differentiate (sin x)^3 directly because sin x's derivative, -cos x, isn't "sitting outside the brackets." (There are, however, other ways to differentiate this, which I will talk about later.) This is where I always screw up because I always divide by the integral by accident, which means that I get the positive/negative signs all wrong.

Oh, and don't forget the...

+ c

If you don't have the derivative sitting outside the brackets... (for trig functions only)

If you have sin x or cos x raised to an odd power, you can do a dirty trick as follows...

If you have, say, (sin x)^3, you can split that up into (sin x)(sin x)^2. Now, (sin x)^2 is also equal to 1 - (cos x)^2. This gives you (sin x)(1 - (cos x)^2), which then gives you sin x - sin x(cos x)^2 which can be integrated to -cos x + (1/3)(cos x)^3 + c!

(And yes, while I was working that one out I made my old mistakes of dividing by the integral instead of the derivative and forgetting the plus c... hopefully I don't make those mistakes in tests...)

If you have sin x or cos x raised to an even power, well, there's a dirty trick for that too, but it's not as convenient.

Here are two more trig identities to memorise (I hate memorisation, so that's just another reason why I hate integrating these things). (cos x)^2 = ((1 + cos 2x)/2) and (sin x)^2 = ((1 - cos 2x)/2). You can go ahead and prove them if you want to feel confident that they work. These functions can be integrated.

If you have (cos x)^6 or whatever, (cos x)^6 = ((cos x)^2)^3 = ((1 + cos 2x)/2)^3. Then you have to do the fun, fun work of expanding that and integrating it. And so on. Argh. I hate these things. I've pretty much said everything about them that I have to, though. Now for some more posts on integration by substitution, differential equations and maybe a quickie on area under a curve as well...