Given: $I=\int \large\frac{\sin 2x }{a \cos ^2 x + b \sin ^2 x }$$dx$
Let $ a\cos ^2 x + b \sin ^2 x=t$
differentiating both the sides
$\Rightarrow\:(-2a\:cosx sinx+2b\:sinxcosx)dx=dt$
$\Rightarrow\:2sinxcosx.(b-a)dx=dt$
But $2sinxcosx=sin2x$
$\therefore\:(b-a).sin2x\:dx=dt$
$\Rightarrow\:sin2x\:dx=\large\frac{1}{b-a}$$.dt$
Substituting the values of $t$ and $dx$ in $I$ we get
$\Rightarrow\:I=\int \large\frac{1}{(b-a)} \times \frac{1}{t} $$.dt$
$\Rightarrow\:I=\large\frac{1}{(b-a)}$$\int \large\frac{1}{t}$$dt=\large\frac{1}{b-a}$$log_e|t|+c$
Since $\int \large\frac{1}{t}$$.dt=log_e |t|$
$\Rightarrow\:I=\large\frac{1}{b-a}$$ \log _e |(a \cos ^2 x + b \sin ^2 x)|+c$
Hence (a) is the correct answer.