# That pesky constant

I've got to work out: $\int \cosec^2(x) \cot(x) \d x$. I did it letting $u = \cosec(x)$ and got an answer -- but when I did it with $u = \cot(x)$, I got something else. What gives?

Ah! A substitution question! My favourite -- and it sounds like you're off to a good start.

If $u = \cosec(x)$, then $\d u = - \cosec(x)\cot(x) \d x$, the integral turns into $- \int u \d u$, which is $C-\frac 12 \cosec^2(x)$.1

How about with $u = \cot(x)$?

In that case, $\d u = -\cosec^2(x) \d x$, so the integral becomes $-\int u \d u$ again, which is $C - \frac{1}{2} \cot^2 (x)$.

### But wait! Those are different answers!

They certainly look that way, don't they? However, I've done something a bit underhand in my answers, which will become apparent.

There's a relationship between $\cot(x)$ and $\cosec(x)$: $1 + \cot^2(x) \equiv \cosec(x)$. Playing around with our second answer, that could be written as $C - \frac{1}{2} \left(\cosec^2 (x) - 1 \right)$, or $(C+\frac{1}{2})- \frac{1}{2} \cosec^2(x)$.

That's very similar to the first answer, but there's a rogue half in there. How do we get rid of that?

Easy. $C$ is an arbitrary constant, which means we could replace $C+\frac12$ with, say, $K$, and everything would be sweet and dandy - we'd have the same answer.

### So, what did I do wrong?

The thing I did wrong was to use the same $C$ for both integrals! There's no reason to expect two arbitrary constants to be the same, and I should have used $C_1$ and $C_2$ or different letters. Using the same letter to mean different things is an easy path to confusion.

* Edited 2014-11-15 to fix some LaTeX

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

1. The constant can go at the end if you prefer, obviously. []

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