Question

If $\int \cos x.\cos 2x.\cos 5xdx=A\sin 2x+B\sin 4x+C\sin 6x+D\sin 8x+k$ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B}+\frac{1}{C}=$

• A. $\frac{1}{A}-\frac{1}{D}$
• B. $\frac{1}{A}+\frac{1}{D}$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

Given, $\int \cos x\cdot \cos 2x\cdot \cos 5xdx$
$=\frac{1}{2}\int 2\cos x\cos 5x\cos 2xdx$
$=\frac{1}{2}\int \{\cos (5x+x)+\cos (5x-x)\}\cos 2xdx$
$=\frac{1}{2}\int (\cos 6x+\cos 4x)\cos 2xdx$
$=\frac{1}{4}\int (2\cos 6x\cos 2x+2\cos 2x\cos 4x)dx$
$=\frac{1}{4}\int (\cos 8x+\cos 4x+\cos 6x+\cos 2x)dx$
$=\frac{1}{4}\left[\frac{\sin 8x}{8}+\frac{\sin 4x}{4}+\frac{\sin 6x}{6}+\frac{\sin 2x}{2}\right]+k$
$=\frac{\sin 2x}{8}+\frac{\sin 4x}{16}+\frac{\sin 6x}{24}+\frac{\sin 8x}{32}+k$
On comparing,
$A=\frac{1}{8},B=\frac{1}{16},C=\frac{1}{24}$ and $D=\frac{1}{32}$
$\therefore \frac{1}{B}+\frac{1}{C}=16+24=40$
Now, $\frac{1}{A}+\frac{1}{D}=8+32=40$
$\therefore \frac{1}{B}+\frac{1}{C}=\frac{1}{A}+\frac{1}{D}$