Question

${\int }^{}\frac{\sin 2x}{\left(a^2{\sin }^{2}x+b^2{\cos }^{2}x\right)}dx$

Answer 1

Put $t=a^2{\sin }^{2}x+b^2{\cos }^{2}x$

$\Rightarrow dt=(2a^2\sin x\cos x-2b^2\cos x\sin x)dx$

$\Rightarrow dt=(a^2\sin 2x-b^2\sin 2x)dx$

$\Rightarrow dt=(a^2-b^2)\sin 2xdx$

$\Rightarrow \frac{dt}{a^2-b^2}=\sin 2xdx$

Now,$\int \frac{\sin 2xdx}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}$

$=\frac{1}{a^2-b^2}\int \frac{dt}{t}$

$=\frac{1}{a^2-b^2}\log \left|t\right|+c$

$=\frac{1}{a^2-b^2}\log |a^2{\sin }^{2}x+b^2{\cos }^{2}x|+c$