Ask Uncle Colin: An Integral The Tutor Couldn't Do

Dear Uncle Colin,

On a recent revision course, my tutor couldn’t integrate ${\int }_0^{\frac{\pi }{2}}\frac{\sin (2\theta )}{1+\cos (\theta )}\mathrm{d}\theta$. 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 $\frac{du}{d\theta }=-\sin (\theta )$, and the limits become $2$ at the bottom and $1$ at the top.

We now have ${\int }_2^1\frac{2(u-1)}{u}(-\frac{du}{d\theta })\mathrm{d}\theta$, which I’d rewrite as

${\int }_1^2\frac{2(u-1)}{u}\mathrm{d}u$.

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

${[2u-2\ln (u)]}_1^2=2-2\ln (2)$.

Hope that helps!

- Uncle Colin