Question

Evaluate $\int cosx\cos 2x\cos 3xdx$

• A. $\frac{1}{48}[2\sin 6x+3\sin 4x+6\sin 2x+12x]+c$
• B. $\frac{1}{48}[2\sin 6x+3sinx+6\sin 2x]+c$
• C. $-\frac{1}{48}[2\sin 6x+3\sin 4x+6sinx+x]+c$
• D. $-\frac{1}{48}[2\sin 6x+3sinx+6\sin 2x]+c$

Answer 1

The correct option is A $\frac{1}{48}[2\sin 6x+3\sin 4x+6\sin 2x+12x]+c$

$cosxcos2xcos3x=\frac{1}{2}\left[2cosxcos2xcos3x\right]$

$=\frac{cos2x}{2}\left[2cosxcos3x\right]$

$=\frac{cos2x}{2}[cos4x+cos2x]$ .....$[2\cos A\cos B=\cos (A+B)+\cos (A-B\left)\right]$

$=\frac{1}{4}\left[2cos4xcos2x\right]+\frac{1}{2}co{s}^{2}2x$

$=\frac{1}{4}[cos6x+cos2x]+\left[\frac{1+cos4x}{4}\right]$

$t\left(x\right)=\frac{cos2x+cos4x+cos6x+1}{4}$

Now, integrating w.r.t. $x$.

$\int t\left(x\right)=\int \frac{cos2x+cos4x+cos6x+1}{4}dx$

$\int t\left(x\right)=\frac{1}{4}[\frac{sin2x}{2}+\frac{sin4x}{4}+\frac{sinx}{6}+x]$

=$\frac{1}{8}[sin2x+\frac{sin4x}{2}+\frac{sin6x}{3}+2x]$

=$\frac{1}{48}[2sin6x+3sin4x+6sin2x+12x]+c$