ok i have a problem and i dont remember how to integrate...
1/(cos^2(u) + 3)du
the function should read 1 over the function of cosine squared of u +3
WolframAlpha to the rescue. Click on Show Steps to see how they got it.
The WolframAlpha shows not a very elegant result for this example
$\displaystyle \displaystyle{\begin{aligned}\int\dfrac{du}{\cos^2 u+3}&=\int\dfrac{du}{\cos^2u+3(\sin^2u+\cos^2u)}=\ int\dfrac{du}{4\cos^2u+3\sin^2u}\\&=\dfrac{1}{4}\i nt\dfrac{1}{1+{\left(\dfrac{\sqrt3}{2}\tan{u}\righ t)\!}^2}\dfrac{du}{\cos^2u}=\left\{\begin{gathered }\dfrac{\sqrt3}{2}\tanu=t\hfill\\\dfrac{du}{\cos^2 u}=\dfrac{2t}{\sqrt3}\,dt\hfill\\\end{gathered}\ri ght\}\\&=\dfrac{1}{2\sqrt3}\int\dfrac{dt}{1+t^2}=\ dfrac{\sqrt3}{6}\arctan{t}+C\\&=\dfrac{\sqrt3}{6}\ arctan\!\left(\dfrac{\sqrt3}{2}\tan{u}\right)+C\en d{aligned}}$