Ask Uncle Colin: an integral that’s giving me a headache

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

Dear Uncle Colin,

I’ve been trying to work out $I = \int_0^{\frac \pi 4} x \frac{\sin(x)}{\cos^3(x)} \d x$ for hours. It’s the fifth time this week I’ve been up until the small hours working on integration and it’s affecting my work and home life. I’m worried I’m becoming a calcoholic.

Books Obdurately Open Late Evening

Dear BOOLE — it’s important to be careful about how much calculus you do, especially if you’re deriving.

However, I can help you with this particular puzzle. Whenever you see the pattern “$x$ multiplied by stuff” under an integral, especially if there’s no clear substitution available, you should immediately think of integration by parts.

In this case, $u$ is going to be $x$, do $\diff u x = 1$. That leaves $\diff v x = \frac{\sin(x)}{\cos^3(x)}$, which looks ugly at first. But not after enough calculus! If you use the substitution $U = \cos(x)$1, you wind up with $v = \frac 1 2 \left(\cos(x)\right)^{-2}$.

Applying the formula, you get:

$I = uv – \int v \diff u x \d x$

$I = \frac 1 2 x\left(\cos(x)\right)^{-2} -\int \frac 1 2 \left(\cos(x)\right)^{-2} \d x$

However, $(\cos(x))^{-2} \equiv \sec^2(x)$, which integrates to $\tan(x)$! Applying the limits:

$I = \left[\frac 1 2 x \sec^2(x) – \frac 1 2 \tan(x)\right]_0^{\frac \pi 4}$

$I = \left( (\frac 1 2)( \frac {\pi}{4}) (2) – \frac 1 2 (1)\right) – \left(\frac{1}{2} (0)(1) -\frac{1}{2}(0)\right)$

$I = \left ( \frac \pi 4 – \frac 12 \right)$

There we go! A nice, sensible answer. When it comes to integrals, though, mine’s a double.

— Uncle 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. carefully using a capital $U$ so as not to confuse it with the other $u$ []


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