Question

Solve:

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

Answer 1

$I=\int \frac{\sin 2x}{a{\cos }^{2}x+b{\sin }^{2}x}dx$

$=\int \frac{\sin 2x}{a-a{\sin }^{2}x+b{\sin }^{2}x}dx$

$=\int \frac{\sin 2x}{a-(a-b){\sin }^{2}x}dx$

$=\frac{1}{b}\int \frac{\sin 2x}{\frac{a}{a-b}-a{\sin }^{2}x}dx$

Let ${\sin }^{2}x=t$ $\Rightarrow 2\sin x.\cos x.dx=dt$

$\Rightarrow \sin 2x.dx=dt$

Then, we have

$I=\frac{1}{a-b}\int \frac{1}{{\left(\sqrt{\frac{a}{a-b}}\right)}^2-(\sqrt{t}{)}^2}.dt$

$=\frac{1}{a-b}.\frac{1}{2\sqrt{\frac{a}{a-b}}}\log \left|\frac{\sqrt{\frac{a}{a-b}}-x}{\sqrt{\frac{a}{a-b}}+x}\right|$

$=\frac{1}{2\sqrt{a(a-b)}}\log \left|\frac{\sqrt{a}-x.\sqrt{a-b}}{\sqrt{a}+x\sqrt{a-b}}\right|+c$