Question

If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k$ (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)

$\frac{1}{A} + \frac{1}{D }$

Answer 2

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