Write a value of $\int \frac{\sin 2 x}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$
We know that $\cos ^{2} x=1-\sin ^{2} x$
$\left(a^{2} \sin ^{2} x+b^{2} \cos ^{2} x\right)=a^{2} \sin ^{2} x+b^{2}\left(1-\sin ^{2} x\right)$
$=\left(a^{2}-b^{2}\right) \sin ^{2} x+b^{2}$
$y=\int \frac{\sin 2 x}{\left(a^{2}-b^{2}\right)(\sin x)^{2}+b^{2}} d x$
Let, $\sin ^{2} x=t$
Differentiating both sides with respect to $x$
$\frac{d t}{d x}=2 \sin x \cos x$
$=\sin 2 x$
$\Rightarrow d t=\sin 2 x d x$
$y=\int \frac{d t}{\left(a^{2}-b^{2}\right) t+b^{2}}$
Use formula $\int \frac{1}{c t+d} d t=\frac{\log (c t+d)}{c}$
$y=\frac{\log \left[\left(a^{2}-b^{2}\right) t+b^{2}\right]}{\left(a^{2}-b^{2}\right)}+c$
Again, put $t=\sin ^{2} x$
$y=\frac{\log \left[\left(a^{2}-b^{2}\right)(\sin x)^{2}+b^{2}\right]}{\left(a^{2}-b^{2}\right)}+c$