Integral $\int \frac{dx}{\sin^2 (x) + \sin(2x)}$ We have to do the following integration 
$$\int \frac{dx}{\sin^2 (x) + \sin(2x)}$$
I tried as
In the numerator, $1 =  \sin^2(x)+\sin(2x)+\cos^2(x)-\sin(2x)$
But now how to proceed?
 A: \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

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

A: HINT:
$$\dfrac1{\sin^2x+\sin2x}=\dfrac{\sec^2x}{\tan^2x+2\tan x}$$
A: 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. 
A: First use the fact that $\sin(2x) = 2\sin(x)\cos(x)$ to obtain
$$\int \frac{1}{\sin^2(x) + 2\cos(x)\sin(x)}dx$$
Divide numerator and denominator by $\sin^2(x)$:
$$\int \frac{\frac{1}{\sin^2(x)}}{1 + 2\frac{\cos(x)}{\sin(x)}} = \int \frac{\csc^2(x)}{1 + 2\cot(x)}dx$$
Then you can use the substitution $u = 2\cot(x)+1$; $du = -2\csc(x)dx$:
$$-\frac{1}{2}\int\frac{1}{u}du = -\frac{\ln(u)}{2}+c$$
And substitute $u$ for $2\cot(x)+1$ again.
