I'm looking at this integral as part of a problem set for a class. So far, we've been given two hints: we should use $x=\arctan(t)$ and we should try dividing the numerator and denominator by $\cos(x)$ to simplify the problem. But even so, it's still pretty tough.
This is the integral: $$\int\frac{\cos^2(x)}{\bigl(\sin^2(x)+4\cos^2(x)\bigr)^2} \, dx $$
On
$$\eqalign{ & I = \int {\frac{{{{\cos }^2}\left( x \right)}}{{{{\left( {{{\sin }^2}\left( x \right) + 4{{\cos }^2}\left( x \right)} \right)}^2}}}dx} = \int {\frac{{{{\cos }^2}\left( x \right)}}{{{{\cos }^4}\left( x \right){{\left( {{{\tan }^2}\left( x \right) + 4} \right)}^2}}}dx} \cr & = \int {\frac{1}{{{{\cos }^2}\left( x \right){{\left( {{{\tan }^2}\left( x \right) + 4} \right)}^2}}}dx} = \int {\frac{1}{{{{\left( {{{\tan }^2}\left( x \right) + 4} \right)}^2}}}d\left( {\tan \left( x \right)} \right)} \cr & {\text{Let}}:t = \tan \left( x \right) \Rightarrow I = \int {\frac{1}{{{{\left( {{t^2} + 4} \right)}^2}}}dt} = \frac{1}{8}\int {\left( {\frac{1}{{{t^2} + 4}} - \frac{{{t^2} - 4}}{{{{\left( {{t^2} + 4} \right)}^2}}}} \right)dt} \cr & = \frac{1}{{16}}\arctan \left( {\frac{t}{2}} \right) - \frac{1}{8}\int {\frac{{{t^2} - 4}}{{{{\left( {{t^2} + 4} \right)}^2}}}dt} = \frac{1}{{16}}\arctan \left( {\frac{t}{2}} \right) + \frac{1}{8}\int {\frac{{\left( {{t^2} + 4} \right) - 2{t^2}}}{{{{\left( {{t^2} + 4} \right)}^2}}}dt} \cr & = \frac{1}{{16}}\arctan \left( {\frac{t}{2}} \right) + \frac{1}{8}\int {\frac{{t'\left( {{t^2} + 4} \right) - t\left( {{t^2} + 4} \right)'}}{{{{\left( {{t^2} + 4} \right)}^2}}}dt} = \frac{1}{{16}}\arctan \left( {\frac{t}{2}} \right) + \frac{1}{8}\frac{t}{{{t^2} + 4}} + C \cr & = \frac{1}{{16}}\arctan \left( {\frac{{\tan \left( x \right)}}{2}} \right) + \frac{1}{8}\frac{{\tan \left( x \right)}}{{{{\tan }^2}\left( x \right) + 4}} + C \cr} $$
Hint: $$ \frac{\;\frac{1}{\cos^4 x}\;}{\frac{1}{\cos^4 x}} \cdot \frac{\cos^2 x}{( \sin^2 x + 4\cos^2 x )^2} = \frac{\sec^2 x}{(\tan^2 x + 4)^2} $$