TITLE: Show that $\int_0^{\pi/3} \big((\sqrt{3}\cos x-\sin x)\sin x\big)^{1/2}\cos x \,dx =\frac{\pi\sqrt{3}}{8\sqrt{2}}. $
QUESTION [19 upvotes]: I have run a FORTRAN code and I have obtained strong evidence that
$$\int_0^{\pi/3} \!\!
\big((\sqrt{3}\cos\vartheta-\sin\vartheta)\sin\vartheta\big)^{\!1/2}\!\cos\vartheta \,d\vartheta
=\frac{\pi\sqrt{3}}{8\sqrt{2}}. 
$$
In fact, it looks like my numerical integration method (the trapezoid rule) converges to the value $\dfrac{\pi\sqrt{3}}{8\sqrt{2}}$, with at least $12$ significant digits.
Any ideas how to prove this result analytically?

REPLY [21 votes]: First, an identity:
$$
\begin{align}
(\sqrt{3}\cos(\theta)-\sin(\theta))\sin(\theta)
&=2\sin(\tfrac\pi3-\theta)\sin(\theta)\\
&=\cos(\tfrac\pi3-2\theta)-\cos(\tfrac\pi3)\\
&=\tfrac12-2\sin^2(\tfrac\pi6-\theta)\\
&=\tfrac12-2\sin^2(\theta-\tfrac\pi6)\tag{1}
\end{align}
$$
Now, the integral
$$
\begin{align}
&\int_0^{\pi/3}
\left((\sqrt{3}\cos(\theta)-\sin(\theta))\sin(\theta)\right)^{1/2}\cos(\theta)\,\mathrm{d}\theta\tag{2}\\
&=\int_0^{\pi/3}
\left(\tfrac12-2\sin^2(\theta-\tfrac\pi6)\right)^{1/2}\cos(\theta)\,\mathrm{d}\theta\tag{3}\\
&=\frac1{\sqrt2}\int_{-\pi/6}^{\pi/6}
\left(1-4\sin^2(\theta)\right)^{1/2}\cos(\theta+\tfrac\pi6)\,\mathrm{d}\theta\tag{4}\\
&=\frac1{\sqrt2}\int_{-\pi/6}^{\pi/6}
\left(1-4\sin^2(\theta)\right)^{1/2}\left(\tfrac{\sqrt3}{2}\cos(\theta)-\tfrac12\sin(\theta)\right)\,\mathrm{d}\theta\tag{5}\\
&=\frac{\sqrt3}{2\sqrt2}\int_{-\pi/6}^{\pi/6}
\left(1-4\sin^2(\theta)\right)^{1/2}\cos(\theta)\,\mathrm{d}\theta\tag{6}\\
&=\frac{\sqrt3}{2\sqrt2}\int_{-\pi/6}^{\pi/6}
\left(1-4\sin^2(\theta)\right)^{1/2}\,\mathrm{d}\sin(\theta)\tag{7}\\
&=\frac{\sqrt3}{4\sqrt2}\int_{-\pi/2}^{\pi/2}
\left(1-\sin^2(\theta)\right)^{1/2}\,\mathrm{d}\sin(\theta)\tag{8}\\
&=\frac{\sqrt3}{4\sqrt2}\int_{-\pi/2}^{\pi/2}
\cos^2(\theta)\,\mathrm{d}\theta\tag{9}\\
&=\frac{\pi\sqrt3}{8\sqrt2}\tag{10}
\end{align}
$$
Explanation:
$\ \:(3)$: Apply $(1)$
$\ \:(4)$: subsitute $\theta\mapsto\theta+\frac\pi6$
$\ \:(5)$: cosine of a sum formula
$\ \:(6)$: toss the odd part over a symmetric domain
$\ \:(7)$: $\mathrm{d}\sin(\theta)=\cos(\theta)\,\mathrm{d}\theta$
$\ \:(8)$: subsitute $\sin(\theta)\mapsto\frac12\sin(\theta)$
$\ \:(9)$: $\sqrt{1-\sin^2(\theta)}=\cos(\theta)$ and $\mathrm{d}\sin(\theta)=\cos(\theta)\,\mathrm{d}\theta$
$(10)$: $\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\theta=\frac\pi2$