Dear Uncle Colin,

On a recent revision course, my tutor couldn’t integrate $\int_0^{\piby2} \frac{\sin(2\theta)}{1+\cos(\theta)} \d \theta$. Can you?

- Reasonable Expectation For University’s Nameless Don?

Hi, REFUND!

Far be it from me to disparage a fellow professional’s integration skills, especially in the heat of the moment; I frequently mistake $2\times 3$ for 5, so I’m not going to cast aspersions.

Fortunately, I can do this, and it stems from spotting something: the top of the fraction can be written as $2\sin(\theta)\cos(\theta)$. And when you have quite so many $\cos$s knocking around with a single $\sin$ on top, it’s screaming out for a cosine substitution.

The ugliest thing here is $1 + \cos(\theta)$ on the bottom, so let’s set that as our $u$.

Then $\diff u \theta = - \sin(\theta)$, and the limits become $2$ at the bottom and $1$ at the top.

We now have $\int_2^1 \frac{2(u-1)}{u} \left(-\diff u \theta \right) \d \theta$, which I’d rewrite as

$\int_1^2 \frac{2(u-1)}{u} \d u$.

Then we can split the fraction up as $2 - \frac{2}{u}$, which integrates very nicely indeed:

$\left[ 2u - 2\ln(u)\right]_1^2 = 2 - 2\ln(2)$.

Hope that helps!

- Uncle Colin