Question

$\int \cos x\cos 2xdx$ is equal to (where $C$ is the constant of integration)

• A. $\frac{1}{2}(\frac{\sin 3x}{3}+\sin x)+C$
• B. $\frac{1}{3}(\frac{\sin 3x}{3}+\sin x)+C$
• C. $\frac{1}{2}(\frac{\sin 2x}{2}+\sin x)+C$
• D. $\frac{1}{2}(\frac{\sin 4x}{4}+\sin x)+C$

Answer 1

The correct option is A $\frac{1}{2}(\frac{\sin 3x}{3}+\sin x)+C$

$\int \cos x\cos 2xdx$
$=\frac{1}{2}\int 2\cos x\cos 2xdx$
$\{\because 2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\right\}$
$=\frac{1}{2}\int (\cos 3x+\cos x)dx=\frac{1}{2}(\frac{\sin 3x}{3}+\sin x)+C$