Show that $\int_0^{\pi/6} \frac{\tan(t)}{\sqrt{\cos(2t)}}dt = \frac{\sin^{-1}1/3}{2}$

Question

I am aware of half-angle identities as well as the identity $\cos (2t)=\cos^{2}t-\sin^{2}t$ but I'm quite lost on how to proceed.

Answer 1

The substitutions $t=\frac{s}{2}$ and $u=\tan(\frac{s}{2})$ simplify the problem. \begin{gather*} \int_{0}^{\pi/6}\dfrac{\tan(t)}{\sqrt{\cos(2t)}}\,\mathrm{d}t=\dfrac{1}{2} \int_{0}^{\pi/3}\dfrac{\tan(\frac{s}{2})}{\sqrt{\cos(s)}}\,\mathrm{d}s = \\[2ex] \dfrac{1}{2} \int_{0}^{1/\sqrt{3}}\dfrac{u}{\sqrt{\frac{1-u^2}{1+u^2}}}\cdot\dfrac{2}{1+u^2}\,\mathrm{d}u = \int_{0}^{1/\sqrt{3}}\dfrac{u}{\sqrt{1-u^4}}\,\mathrm{d}u= \\[2ex] \dfrac{1}{2}\left[\arcsin(u^2)\right]_{0}^{1/\sqrt{3}} = \dfrac{1}{2}\arcsin(1/3). \end{gather*}