The value of $\int \frac{\sin x}{\sin 4 x} d x$ is

Moderate

A

$\frac{1}{4} \log |\frac{\sin x - 1}{\sin x + 1}| - \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} \sin x - 1}{\sqrt{2} \sin x + 1}| + C$
B

$\frac{1}{8} \log |\frac{\cos x - 1}{\cos x + 1}| - \frac{1}{2 \sqrt{2}} \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + C$
C

$\frac{1}{8} \log |\frac{\sin x - 1}{\sin x + 1}| - \frac{1}{4 \sqrt{2}} \log |\frac{\sqrt{2} \sin x - 1}{\sqrt{2} \sin x + 1}| + C$
D

none of these

Solution

The given integrand can be written as
$\frac{1}{4 \cos x \cos 2 x} = \frac{\cos x}{4 (1 - \sin^{2} x) (1 - 2 \sin^{2} x)}$ , hence
$\begin{array}{l} \int \frac{\sin x}{\sin 4 x} d x = \frac{1}{4} \int \frac{d t}{(1 - t^{2}) (1 - 2 t^{2})} (t = \sin x) \\ = \frac{1}{4} \int [\frac{1}{t^{2} - 1} - \frac{1}{t^{2} - 1 / 2}] d t \\ = \frac{1}{4} [\frac{1}{2} \log |\frac{t - 1}{t + 1}| - \frac{1}{\sqrt{2}} \log |\frac{t - 1 / \sqrt{2}}{t + 1 / \sqrt{2}}|] + C \\ = \frac{1}{8} \log |\frac{\sin x - 1}{\sin x + 1}| - \frac{1}{4 \sqrt{2}} \log |\frac{\sqrt{2} \sin x - 1}{\sqrt{2} \sin x + 1}| + C . \end{array}$

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