Question

\[\int\text{ cos x  cos  2x   cos  3x  dx}\]

Answer 1

\[\int\text{ cos x . cos  2x .   cos  3x   dx}\]
\[ \Rightarrow \frac{1}{2}\int\left[ 2 \cos 2x \cdot \cos x \right] \text{ cos  3x  dx}\]
\[ \Rightarrow \frac{1}{2}\int\left[ \text{ cos } \left( 2x + x \right) + \text{ cos } \left( 2x - x \right) \right] \text{ cos  3x  dx} ..............\left[ \because 2\text{ cos }A\text{ cos B }= \cos \left( A + B \right) + \text{ cos }\left( A - B \right) \right]\]
\[ \Rightarrow \frac{1}{2}\int\left( \cos3x + \cos x \right) \text{ cos  3x dx }\]
\[ \Rightarrow \frac{1}{2}\int \text{ cos }^2 \text{ 3x  dx} + \frac{1}{2}\int\text{ cos } 3x \cdot \text{ cos  x dx}\]
\[ \Rightarrow \frac{1}{2}\int\left( \frac{1 + \text{ cos }6x}{2} \right)dx + \frac{1}{4}\int2 \text{ cos  3x} \cdot \text{ cos x dx} ...................\left[ \because \cos 2x = \cos^2 x - 1 \right]\]
\[ \Rightarrow \frac{1}{4}\left[ x + \frac{\sin 6x}{6} \right] + \frac{1}{4}\int\left( \cos 4x + \cos 2x \right)dx\]
\[ \Rightarrow \frac{1}{4}\left[ x + \frac{\sin 6x}{6} \right] + \frac{1}{4}\left[ \frac{\sin 4x}{4} + \frac{\sin 2x}{2} \right] + C\]
\[ \Rightarrow \frac{x}{4} + \frac{\sin 6x}{24} + \frac{\sin 4x}{16} + \frac{\sin 2x}{8} + C\]