Problem source

Show that
\[
\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \frac 1{1-\cos2\theta} \;\d\theta = \frac{\sqrt3}2 - \frac12\;.
\]
By using the substitution $x=\sin2\theta$, or otherwise, show that
\[
\int_{\sqrt3/2}^1 \frac 1 {1-\sqrt{1-x^2}} \, \d x = \sqrt 3 -1 -\frac\pi 6 \;.
\]
Hence evaluate the integral
\[
\int_1^{2/\sqrt3} \frac  1{y ( y - \sqrt{y^2-1^2})} \, \d y \;.
\]

Solution source

\begin{align*}
&& I &= \int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \frac 1{1-\cos2\theta} \;\d\theta \\
&&&= \int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \frac 1{2\sin^2 \theta} \;\d\theta \\
&&&= \frac12\int_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \cosec^2 \theta \;\d\theta \\
&&&= \frac12\left [-\cot \theta \right]_{\frac{1}{6}\pi}^{\frac{1}{4}\pi} \\
&&&= \frac12 \left (\cot \frac{\pi}{6} - \cot \frac{\pi}{4} \right)\\
&&&= \frac{\sqrt{3} - 1}{2}
\end{align*}

\begin{align*}
&& J &= \int_{\sqrt3/2}^1 \frac 1 {1-\sqrt{1-x^2}} \, \d x \\
x = \sin 2 \theta, \d x = 2\cos 2\theta \d \theta &&&=\int_{\pi/6}^{\pi/4} \frac{2 \cos 2 \theta }{1-\cos 2 \theta} \d \theta \\
&&&=\int_{\pi/6}^{\pi/4} \frac{2 \cos 2 \theta -2+2}{1-\cos 2 \theta} \d \theta \\
&&&= -2\left (\frac{\pi}{4} - \frac{\pi}6 \right) + 2I \\
&&&= \sqrt{3}-1-\frac{\pi}{6}
\end{align*}

\begin{align*}
&& K &= \int_1^{2/\sqrt{3}} \frac{1}{y(y-\sqrt{y^2-1})} \d y \\
y = 1/x, \d x = -1/y^2 \d y &&&= \int_{x=1}^{x=\sqrt{3}/2} \frac{1}{1-\sqrt{1-x^2}} \d x\\
&&&= \sqrt{3}-1 -\frac{\pi}6
\end{align*}