# Integral &#x222B;<!-- ∫ --> d x </mrow> sin

Integral $\int \frac{dx}{{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(2x\right)}$
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kejohananws
$\begin{array}{rl}\int \frac{dx}{{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(2x\right)}& =\int \frac{{\mathrm{sec}}^{2}\left(x\right)}{{\mathrm{sin}}^{2}\left(x\right){\mathrm{sec}}^{2}\left(x\right)+2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right){\mathrm{sec}}^{2}\left(x\right)}dx\\ & =\int \frac{{\mathrm{sec}}^{2}\left(x\right)}{{\mathrm{tan}}^{2}\left(x\right)+2\mathrm{tan}\left(x\right)}dx\end{array}$
Letting $u:=\mathrm{tan}\left(x\right)$, then $du={\mathrm{sec}}^{2}\left(x\right)dx$ gives
$\begin{array}{rl}\int \frac{dx}{{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(2x\right)}& =\int \frac{du}{{u}^{2}+2u}\\ & =\int \frac{du}{\left(u+1{\right)}^{2}-1}\end{array}$
Letting z:=u+1, then dz=du gives

Reersing the final substitution gives