Question

Solve $\int \cos 2x\cos 4xcos6xdx$.

• A. $I=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}-\frac{\sin 4x}{4}+x]+c$
• B. $I=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+c$
• C. $I=\frac{5}{4}[\frac{\sin 12x}{12}-\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+c$
• D. None of these

Answer 1

The correct option is B $I=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+c$

We know:- $\cos A\cos B=1/2\left[\cos \right(A+B)+\cos (A-B\left)\right]$

now, $\int \left(\cos 2x\cos 4x\cos 6x\right).dx$

$\int cos2x\left[\frac{1}{2}\right(cos10x+cos2x\left)\right]dx$

$=\int \frac{1}{2}(cos2xcos10x+co{s}^{2}2x)dx$

$=\int \frac{1}{4}[cos12x+cos8x]dx+\int \frac{1}{4}(1+cos4x)dx$

$=\int \frac{1}{4}[\cos 12x+\cos 8x+\cos 4x+1]dx$

$=\frac{1}{4}[\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+x]+c$