Dear Uncle Colin,

I was doing a STEP paper and it asked me to calculate $\int_0^1 x^3 \arctan\left(\frac{1-x}{1+x}\right) \dx$, given that $\int_0^1 \frac{x^4}{1+x^2} \dx = \frac{\pi}{4} - \frac{2}{3}$.


College Asked Me Back: Rocked Interview. Daren’t Get Excited

Hello, CAMBRIDGE! Is this thing on?

Even with the given hint, this is a bit of a mess; however, the $\arctan$ knocking about in there is a dead giveaway. It’s integration by parts, and the inverse trig is your $u$.

In case you didn’t know that $\frac{\d}{\d z} \arctan(z) = \frac{1}{1+z^2}$, you can get it implicitly: if $y = \arctan(z)$, then $\tan(y) = z$; $\sec^2(y) \diff yz = 1$, so $\diff yz = \frac{1}{1 + \tan^2(y)} = \frac{1}{1 + z^2}$.

That’s all well and good, but what about $\frac{\d}{\dx} \arctan\left( \frac{1-x}{1+x} \right)$? Well, that’s just chain rule. The derivative of the argument is $\frac{-2}{(1+x)^2}$, so your $u’$ works out to be $\frac{1}{1 + z^2} \cdot \frac{-2}{(1+x)^2}$, or $\frac{-2}{(1+x)^2 + (1+x)^2z^2}$.

Why have I written it like that? It’s because $(1+x)^2 z^2 = (1-x)^2$, going back to the definition, so the bottom of the fraction is $(1+x)^2 + (1-x)^2$, or $2 + 2x^2$ – giving you $u’ = \frac{-1}{1+x^2}$. This is a good sign: there’s a $1+x^2$ in the hint we’re given.

If $v’=x^3$, then $v = \frac 14 x^4$, just for the sake of completeness.

Now put it all together: the integral is $\left[ uv \right]_0^1 - \int_0^1 vu’ \dx$. That’s $\left[\frac 14 x^4 \arctan\left( \frac{1-x}{1+x} \right) \right]_0^1 + \frac{1}{4} \int_0^1 \frac{x^4}{1+x^2} \dx$ – and lookit, we’ve got something we know about as our last term!

Better yet, when $x=1$, the $\arctan$ vanishes and you’re left with 0 for the big bracket; when $x=0$, the $x^4$ vanishes, so the big bracket can be completely ignored. That leaves you with a quarter of the integral you’re given as a hint, so you end up with $\frac{\pi}{16} - \frac {1}{6}$.

Good luck with the STEP!

-- Uncle Colin