Question

If $\int \cos x\cos 2x\cos 3xdx=A\sin 6x+B\sin 4x+P\sin 2x+Qx+C$
$($where $C$ is constant of integration$),$ then which of the following is/are true:

• A. $\frac{1}{A}+\frac{1}{P}=32$
• B. $\frac{1}{B}-\frac{1}{Q}=12$
• C. $\frac{1}{B}+\frac{1}{P}=24$
• D. $\frac{1}{A}-\frac{1}{Q}=20$

Answer 1

Answer: (A), (B), (C), (D)

$\frac{1}{A}-\frac{1}{Q}=20$

$I=\int \cos x\cos 2x\cos 3xdx=\frac{1}{2}\int \left(2\cos 3x\cos x\right)\cdot \cos 2xdx=\frac{1}{2}\int (\cos 4x+\cos 2x)\cdot \cos 2xdx=\frac{1}{4}(2\cos 4x\cos 2x+2{\cos }^{2}2x)dx=\frac{1}{4}\int (\cos 6x+\cos 2x+1+\cos 4x)dx=\frac{1}{4}[\frac{1}{6}\sin 6x+\frac{1}{4}\sin 4x+\frac{1}{2}\sin 2x+x]+C\therefore A=\frac{1}{24},B=\frac{1}{16},P=\frac{1}{8},Q=\frac{1}{4}\Rightarrow \frac{1}{A}+\frac{1}{P}=24+8=32\frac{1}{B}-\frac{1}{Q}=16-4=12\frac{1}{B}+\frac{1}{P}=16+8=24\frac{1}{A}-\frac{1}{Q}=24-4=20$