# Ask Uncle Colin: A Seemingly Undefined Integral

Dear Uncle Colin,

I need to evaluate $\int_0^{\piby2} \frac{1}{1+\sin(x)}\dx$ but I end up with $\infty - \infty$ and that’s no good! How should I be doing it?

Big Integral, Not Exactly Trivial

Hi, BINET, and thanks for your message!

This is a fun problem! I can think of several possible approaches.

### What I suspect you did

The ‘obvious’ approach here is to multiply top and bottom by $(1-\sin(x))$, making the integral $\int_0^{\piby 2} \frac{1-\sin(x)}{\cos^2(x)}\dx$.

That can be rewritten as $\int_0^{\piby 2} \sec^2(x) - \sec(x)\tan(x) \dx$ and integrated as $\left[ \tan(x) - \sec(x) \right]_0^{\piby 2}$.

However, neither of the terms is defined at $\piby 2$, which is what you get when you divide by something that’s not defined throughout the interval1

### Approach 1: L’Hopital

The resulting integral is $\tan(x) - \sec(x)$, or $\frac{\sin(x) - 1}{\cos(x)}$. The numerator and denominator are both zero at $\piby 2$, so we can apply L’Hopital’s rule and say that the value in the limit is $\frac{\cos(x)}{-\sin(x)}$, which is zero.

The integral is then $[(0) - (-1)] = 1$.

### Approach 2: Symmetry

The original integral, $I= \int_0^{\piby 2} \frac{1}{1+\sin(x)}\dx$, is equal to $\int_{\piby 2}^{\pi} \frac{1}{1+\sin(x)}\dx$, because of symmetry. That means $2I = \int_0^{\pi} \frac{1}{1+\sin(x)}\dx$, or $\left[\tan(x) - \sec(x)\right]_0^\pi = 2$.

Again, $I= 1$.

### Approach 3: Substitution

If you use the substitution $u = \pi - x$ in the original integral, you wind up with $\int_0^{\piby 2} \frac{\cos(u) - 1}{\sin^2(u)} \d{u}$. That integrates to $\left[\cot(u) - \cosec(u)\right]_0^{\piby2}$.

Aha! That’s solved the problem at $\piby 2$! However, that comes at a price: neither term is defined at $u=0$.

However, we have small angle approximations. $\cosec(u)-\cot(u) = \frac{1-\cos(u)}{\sin(u)}$, which - for small $u$ - is very close to $\frac{u^2/2}{u}$, which approaches 0 in the limit. At $u=\piby 2$, the expression is 1, so we end up with 1 again.

I’m sure there are other ideas I haven’t considered! All the same, I hope one of these helps.

- Uncle Colin

## 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. OK, OK, it’s defined in the limit, but it’s still asking for trouble. []

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