Ask Uncle Colin: A STEP integral

Dear Uncle Colin,

I was doing a STEP paper and it asked me to calculate ${\int }_0^1{x}^3\arctan \left(\frac{1-x}{1+x}\right)\mathrm{d}x$, given that ${\int }_0^1\frac{x^4}{1+x^2}\mathrm{d}x=\frac{\pi }{4}-\frac{2}{3}$.

Nut-uh.

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 $\frac{\mathrm{d}}{\mathrm{d}z}\arctan (z)=\frac{1}{1+z^2}$, you can get it implicitly: if $y=\arctan (z)$, then $\tan (y)=z$; ${\sec }^2(y)\frac{dy}{dz}=1$, so $\frac{dy}{dz}=\frac{1}{1+{\tan }^2(y)}=\frac{1}{1+z^2}$.

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

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 $u^{\prime }=\frac{-1}{1+x^2}$. This is a good sign: there’s a $1+x^2$ in the hint we’re given.

If $v^{\prime }=x^3$, then $v=\frac{1}{4}{x}^4$, just for the sake of completeness.

Now put it all together: the integral is ${\left[uv\right]}_0^1-{\int }_0^{1}v{u}^{\prime }\mathrm{d}x$. That’s ${[\frac{1}{4}{x}^4\arctan \left(\frac{1-x}{1+x}\right)]}_0^1+\frac{1}{4}{\int }_0^1\frac{x^4}{1+x^2}\mathrm{d}x$ – 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