$ \text { Let } I =\int \frac{\sin 2 x}{3 \sin ^4 x-4 \sin ^2 x+1} d x$
$=\int \frac{\sin 2 x}{3\left(\sin ^2 x\right)^2-4 \sin ^2 x+1} d x $
Put $\sin ^2 x=t$
$ \therefore 2 \sin x \cos x d x=d t$
$\therefore \sin 2 x d x=d t$
$\therefore I=\int \frac{ dt }{3 t ^2-4 t +1}$
$=\int \frac{ dt }{3\left( t ^2-\frac{4}{3} t +\frac{1}{3}\right)}$
$\left(\frac{1}{2} \text { coefficient of } t\right)^2=\left[\frac{1}{2} \times\left(\frac{-4}{3}\right)\right]^2=\frac{4}{9}$
$\therefore I =\frac{1}{3} \int \frac{1}{ t ^2-\frac{4}{3} t +\frac{4}{9}-\frac{4}{9}+\frac{1}{3}} dt$
$=\frac{1}{3} \int \frac{1}{\left( t ^2-\frac{4}{3} t +\frac{4}{9}\right)-\frac{1}{9}} d t$
$=\frac{1}{3} \int \frac{1}{\left( t -\frac{2}{3}\right)^2-\left(\frac{1}{3}\right)^2} dt$
$=\frac{1}{3} \cdot \frac{1}{2 \times \frac{1}{3}} \log \left|\frac{\left( t -\frac{2}{3}\right)-\frac{1}{3}}{\left( t -\frac{2}{3}\right)+\frac{1}{3}}\right|+ c$
$=\frac{1}{2} \log \left|\frac{3 t -3}{3 t -1}\right|+ c$
$\therefore I=\frac{1}{2} \log \left|\frac{3 \sin ^2 x-3}{3 \sin ^2 x-1}\right|+c$
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