$$
\begin{align}
\int_0^1\sqrt{k^2+\beta^2}\,\mathrm{d}k
&=\beta^2\int_0^{1/\beta}\sqrt{k^2+1}\,\mathrm{d}k\\
&=\beta^2\int_0^{\pi/2-\arctan(\beta)}\sec^3(\theta)\,\mathrm{d}\theta\\
&=\beta^2\int_0^{\pi/2-\arctan(\beta)}\frac1{(1-\sin^2(\theta))^2}\,\mathrm{d}\sin(\theta)\\
&=\beta^2\int_0^{1\big/\sqrt{1+\beta^2}}\frac1{(1-u^2)^2}\,\mathrm{d}u\\
&=\beta^2\int_0^{1\big/\sqrt{1+\beta^2}}\frac14\left(\frac1{(1-u)^2}+\frac1{1-u}+\frac1{(1+u)^2}+\frac1{1+u}\right)\,\mathrm{d}u\\
&=\frac{\beta^2}{4}\left[\frac{2u}{1-u^2}+\log\left(\frac{1+u}{1-u}\right)\right]_0^{1\big/\sqrt{1+\beta^2}}\\

Here is a $11^{\text{th}}$ BMO 1975 question:

Let $m$ be a fixed positive integer. You are given that$$\binom{2m}0+\binom{2m}1\cos\theta+\binom{2m}2\cos2\theta+\dots+\binom{2m}{2m}\cos2m\theta=\left(2\cos\frac12\theta\right)^{2m}\cos \,m\theta$$where there are $2m+1$ terms on the LHS; the value of each side of this identity is defined to be $f_m(\theta)$. The function $\mathrm g_m(\theta)$is defined by$$\mathrm g_m(\theta)=\binom{2m}0+\binom{2m}2\cos2\theta+\binom{2m}4\cos4\theta+\dots+\binom{2m}{2m}\cos2m\theta.$$Given that there is no rational $k$ for which $\alpha = k\pi$, find the valu…