Question

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{6} \cos x \cos 2x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{6} \cos x \left( \cos^2 x - \sin^2 x \right) dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{6} \left( 2 \cos^3 x - \cos x \right) dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{6} \left( 2 \cos x \left( 1 - \sin^2 x \right) - \cos x \right) dx\]
\[ \Rightarrow I = \left[ 2\left( \sin x - \frac{\sin^3 x}{3} \right) - \sin x \right]_0^\frac{\pi}{6} \]
\[ \Rightarrow I = \left[ 2\left( \frac{1}{2} - \frac{1}{24} \right) - \frac{1}{2} \right] - 0\]
\[ \Rightarrow I = \frac{5}{12}\]