$E v a l u a t e \int \frac{\sin x}{\sin 4 x} d x$

Difficult

A

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

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

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

None of these

Solution

$l e t I = \int \frac{\sin x}{\sin 4 x} d x$
$= \int \frac{\sin x}{2 \sin 2 x \cos 2 x} d x$
$= \int \frac{\sin x}{4 \sin x \cos x \cos 2 x} d x$
$= \frac{1}{4} \int \frac{1}{\cos x \cos 2 x} d x$
$= \frac{1}{4} \int \frac{\cos x}{\cos^{2} x \cos 2 x} d x$
$= \frac{1}{4} \int \frac{\cos x}{(1 - \sin^{2} x) (1 - 2 \sin^{2} x)} d x$
$p u t t i n g \sin x = t a n d \cos x d x = d t, w e g e t$
$I = \frac{1}{4} \int \frac{d t}{(1 - t^{2}) (1 - 2 t^{2})}$
$= \frac{1}{4} \int [\frac{2}{1 - 2 t^{2}} - \frac{1}{1 - t^{2}}] d t$
$= - \frac{1}{4} \int \frac{d t}{(1 - t^{2})} + \frac{2}{4} \int \frac{d t}{1 - {(\sqrt{2} t)}^{2}}$
$= - \frac{1}{4} x \frac{1}{2} \log |\frac{1 + t}{1 - t}| + \frac{1}{2} \cdot \frac{1}{2 \sqrt{2}} \log |\frac{1 + \sqrt{2} t}{1 - \sqrt{2} t}| + C$

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

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