Question
$\int \frac{\sin x}{\sin 3 x} d x$
✓
Answer
$ \text { Let } I =\int \frac{\sin x}{\sin 3 x} d x$
$=\int \frac{\sin x}{3 \sin x-4 \sin ^3 x} \cdot d x$
$=\int \frac{\sin x}{\sin x\left(3-4 \sin ^2 x\right)} \cdot d x$
$=\int \frac{1}{3-4 \sin ^2 x} d x $
Dividing numerator and denominator by $\cos ^2 x$, we get
$ I =\int \frac{\sec ^2 x}{3 \sec ^2 x-4 \tan ^2 x} \cdot d x$
$=\int \frac{\sec ^2 x}{3\left(1+\tan ^2 x\right)-4 \tan ^2 x} \cdot d x$
$=\int \frac{\sec ^2 x}{3-\tan ^2 x} d x $
Put $\tan x=t$
$\therefore \sec ^2 x d x=d t$
$\therefore I =\int \frac{ dt }{3- t ^2}$
$=\int \frac{1}{(\sqrt{3})^2- t ^2} dt$
$=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+ t }{\sqrt{3}- t }\right|+ c$
$\therefore I =\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+c I $
$=\int \frac{\sin x}{3 \sin x-4 \sin ^3 x} \cdot d x$
$=\int \frac{\sin x}{\sin x\left(3-4 \sin ^2 x\right)} \cdot d x$
$=\int \frac{1}{3-4 \sin ^2 x} d x $
Dividing numerator and denominator by $\cos ^2 x$, we get
$ I =\int \frac{\sec ^2 x}{3 \sec ^2 x-4 \tan ^2 x} \cdot d x$
$=\int \frac{\sec ^2 x}{3\left(1+\tan ^2 x\right)-4 \tan ^2 x} \cdot d x$
$=\int \frac{\sec ^2 x}{3-\tan ^2 x} d x $
Put $\tan x=t$
$\therefore \sec ^2 x d x=d t$
$\therefore I =\int \frac{ dt }{3- t ^2}$
$=\int \frac{1}{(\sqrt{3})^2- t ^2} dt$
$=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+ t }{\sqrt{3}- t }\right|+ c$
$\therefore I =\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+c I $
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