Nog zo wat...

$
\eqalign{
  & \int {{1 \over {\cos ^6 x}}} \,dx = \int {{1 \over {\cos ^4 x}} \cdot {1 \over {\cos ^2 x}}} \,dx = \int {\left( {{1 \over {\cos ^2 x}}} \right)^2  \cdot {1 \over {\cos ^2 x}}} \,dx  \cr
  & Wat\,\,nu?  \cr
  & \tan ^2 x = {{\sin ^2 x} \over {\cos ^2 x}} = {{1 - \cos ^2 x} \over {\cos ^2 x}} = {1 \over {\cos ^2 x}} - 1 \Rightarrow {1 \over {\cos ^2 x}} = \tan ^2 x + 1  \cr
  & Geeft:  \cr
  & \int {\left( {\tan ^2 x + 1} \right)^2  \cdot } \,d(\tan x)\buildrel {u = \tan x} \over
 \longrightarrow \int {\left( {u^2  + 1} \right)^2 du}  \cr}
$

Zie ook Integralen of Goniometrische integraal berekenen

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