Question

$\int \frac{1}{{\sin }^{2}x+\sin 2x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{{\sin }^{2}x+\sin \left(2x\right)}dx \\ =\int \frac{1}{{\sin }^{2}x+2\sin x\cos x}dx \\ \mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}{\cos }^{2}x \\ \Rightarrow I=\int \frac{{\sec }^{2}x}{{\tan }^{2}x+2\tan x}dx \\ \mathrm{Let}\tan x=t \\ \Rightarrow {\sec }^{2}xdx=dt \\ \therefore I=\int \frac{dt}{t^2+2t} \\ =\int \frac{dt}{t^2+2t+1-1} \\ =\int \frac{dt}{{\left(t+1\right)}^2-{\left(-1\right)}^2} \\ =\frac{1}{2}\ln \left|\frac{t+1-1}{t+1+1}\right|+\mathrm{C} \\ =\frac{1}{2}\ln \left|\frac{t}{t+2}\right|+\mathrm{C} \\ =\frac{1}{2}\ln \left|\frac{\tan x}{\tan x+2}\right|+\mathrm{C} \end{gathered}$$