Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sin 2x}{a{\cos }^{2}x+b{\sin }^{2}x}dx \\ =\int \frac{\sin 2x}{a\left(1-{\sin }^{2}x\right)+b{\sin }^{2}x}dx \\ =\int \frac{\sin 2x}{\left(b-a\right){\sin }^{2}x+a}dx \\ \mathrm{Putting}s{\mathrm{in}}^{2}x=t \\ \Rightarrow 2\sin x.\cos x=\frac{dt}{dx} \\ \Rightarrow \sin 2x=\frac{dt}{dx} \\ \Rightarrow \sin 2xdx=dt \\ \therefore I=\int \frac{1}{\left(b-a\right)t+a}dt \\ =\frac{1}{\left(b-a\right)}\ln \left|\left(b-a\right)t+a\right|+\mathrm{C}\left[\because \int \frac{1}{\mathrm{ax}+b}dx=\frac{1}{a}\ln \left|ax+b\right|+\mathrm{C}\right] \\ =\frac{1}{\left(b-a\right)}\ln \left|\left(b-a\right){\sin }^{2}x+a\right|+\mathrm{C}\left[\because t={\sin }^{2}x\right] \\ =\frac{1}{\left(b-a\right)}\ln \left|b{\sin }^{2}x+a\left(1-{\sin }^{2}x\right)\right|+\mathrm{C} \\ =\frac{1}{\left(b-a\right)}\ln \left|b{\sin }^{2}x+a{\cos }^{2}x\right|+\mathrm{C} \end{gathered}$$