Integral $\int \frac{dx}{\sin^2 (x) + \sin(2x)}$

\begin{align} \int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{\sec^2(x)}{\sin^2(x)\sec^2(x)+2\sin(x)\cos(x)\sec^2(x)}dx\\ &=\int\frac{\sec^2(x)}{\tan^2(x)+2\tan(x)}dx \end{align}

Letting $u:=\tan(x)$, then $du=\sec^2(x)dx$ gives

\begin{align} \int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{du}{u^2+2u}\\ &=\int\frac{du}{(u+1)^2-1} \end{align}

Letting $z:=u+1$, then $dz = du$ gives

\begin{align} \int\frac{dx}{\sin^2(x)+\sin(2x)}&=\int\frac{dz}{z^2-1}\\ &=-\text{artanh }(z)+C\\ &=-\text{artanh }(u+1)+C\\ \end{align}

Reversing the final substitution gives

$$\int\frac{dx}{\sin^2(x)+\sin(2x)} = -\text{artanh }(\tan(x)+1)+C$$

HINT:

$$\dfrac1{\sin^2x+\sin2x}=\dfrac{\sec^2x}{\tan^2x+2\tan x}$$

We can transform our integral as: $$I =\int \frac {1}{\sin^2 x +\sin 2x} dx =\int -(\frac {-1}{2\cot x+1})\mathrm {cosec}^2 x dx $$ Using $v=\cot x $, this can be easily solved. Hope it helps.