$$ \begin{aligned} & \text { If } \int \cos x \cdot \cos 2 x \cdot \cos 5 x d x \\ & =A \sin 2 x+B \sin 4…
$$
\begin{aligned}
& \text { If } \int \cos x \cdot \cos 2 x \cdot \cos 5 x d x \\
& =A \sin 2 x+B \sin 4 x+C \sin 6 x+D \sin 8 x+k
\end{aligned}
$$
(where $k$ is the arbitrary constant of integration), then $\frac{1}{B}+\frac{1}{C}=$
$\frac{1}{A}-\frac{1}{D}$
$\frac{1}{A}+\frac{1}{D}$
1
0
Solution
Given,
$
\begin{aligned}
\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
\end{aligned}
$
On comparing,
$
\begin{aligned}
A & =\frac{1}{8}, B=\frac{1}{16}, C=\frac{1}{24} \text { and } D=\frac{1}{32} \\
\therefore \quad \frac{1}{B}+\frac{1}{C} & =16+24=40
\end{aligned}
$
Now, $\frac{1}{A}+\frac{1}{D}=8+32=40$
$
\therefore \quad \frac{1}{B}+\frac{1}{C}=\frac{1}{A}+\frac{1}{D}
$