Question

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{6} \cos^{- 3} 2\theta \sin\ 2\theta\ d\ \theta . Then, \]

\[I = \int_0^\frac{\pi}{6} \frac{\sin 2\theta}{\cos^3 2\theta} d \theta\]

\[Let\ \cos 2\theta = t . Then, - 2 \sin 2\theta\ d\theta = dt\]

\[When\ \theta = 0, t = 1\ and\ \theta = \frac{\pi}{6}, t = \frac{1}{2}\]

\[ \therefore I = \frac{- 1}{2} \int_1^\frac{1}{2} \frac{dt}{t^3}\]

\[ \Rightarrow I = \frac{1}{2} \left[ \frac{1}{2 t^2} \right]_1^\frac{1}{2} \]

\[ \Rightarrow I = \frac{1}{2}\left( 2 - \frac{1}{2} \right)\]

\[ \Rightarrow I = \frac{3}{4}\]