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Het bepalen van $
\eqalign{\int {\frac{{\cot x}}
{{1 + \sin x}}dx}}
$ lukt me niet. Ik heb de $cot x$ geschreven als $\eqalign{\frac{cosx}{sinx}}$ en daarna substitutie van $u=sin x$
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$ \eqalign{ & \int {\frac{{\cot (x)}} {{1 + \sin (x)}}} \,dx = \cr & \int {\frac{1} {{1 + \sin (x)}} \cdot \frac{{\cos (x)}} {{\sin (x)}}} \,dx = \cr & \int {\frac{1} {{\sin (x)(1 + \sin (x))}}} \,d(\sin (x)) = \cr & \int {\frac{1} {{u(u + 1)}}du = } \cr & \int {\frac{1} {u} - \frac{1} {{u + 1}}\,du = } \cr & \ln (u) - \ln (u + 1) \cr & \ln (\sin (x)) - \ln (\sin (x) + 1) \cr & \ln \left( {\frac{{\sin (x)}} {{\sin (x) + 1}}} \right) \cr & - \ln \left( {\frac{{\sin (x) + 1}} {{\sin (x)}}} \right) \cr & - \ln \left( {1 + \frac{1} {{\sin (x)}}} \right) \cr} $
