I have tried to use trigonometric transformation for coming up to the simple fraction for integration of :$$\int\frac{dx}{\sin^2x-\cos^4x}$$
However, I don't succed , I seek for simple method by part or by fraction or by variable change to do the titled integral ?
On
$$ \eqalign{ & {1 \over {\sin ^{\,2} x - \cos ^{\,4} x}} = {1 \over {1 - \cos ^{\,2} x - \cos ^{\,4} x}} = \cr & = - {1 \over {\left( {\cos ^{\,2} x + \varphi } \right)\left( {\cos ^{\,2} x - 1/\varphi } \right)}} = \cr & = - {4 \over {\left( {\cos (2x) + 1 + 2\varphi } \right)\left( {\cos (2x) + 1 - 2/\varphi } \right)}} = \cr & = {4 \over {2\left( {\varphi + 1/\varphi } \right)}}\left( {{1 \over {\left( {\cos (2x) + 1 + 2\varphi } \right)}} - {1 \over {\left( {\cos (2x) + 1 - 2/\varphi } \right)}}} \right) = \cr & = 2{\varphi \over {\varphi + 2}}\left( {{1 \over {\left( {\cos (2x) + 1 + 2\varphi } \right)}} - {1 \over {\left( {\cos (2x) + 1 - 2/\varphi } \right)}}} \right) = \cr & = \quad \cdots \cr} $$ where $\varphi$ is the golden ratio. Can you continue from here ?
Put $t= \tan x$ then $dt = {1\over \cos^2x}dx = (t^2+1)dx$ and
\begin{eqnarray}\int\frac{dx}{\sin^2x-\cos^4x}&=&\int\frac{\sin^2x+\cos^2x}{\sin^2x-\cos^4x}\,dx\\ &=&\int\frac{\tan^2x+1}{\tan^2x-\cos^2x}\,dx\\&=& \int\frac{\tan^2x+1}{\tan^2x-{1\over \tan^2x+1}}\,dx\\&=& \int\frac{(t^2+1)dt}{t^4+t^2-1}\\&=&...\end{eqnarray}