Hallo,
substituiere z= sin^2(y) +1
allgemein gilt: sin(2y)= 2 sin(y) cos(y) ->dann kürzen
\( z=\sin ^{2}(y)+1 \)
\( \frac{dz}{dy}=2 \sin (y) \cos (y) \)
$$ d y=\frac{d z}{2 \sin (y) \cos (y)} $$
\( =\int \frac{\sin (2 y)}{2 \cdot z} \cdot \frac{d z}{2 \sin (y) \cos (y)} \)
$$ \begin{array}{l} {=\frac{1}{2} \int \frac{1}{z} d z} \\ {=\frac{1}{2} \cdot \ln |z|+c=\frac{1}{2} \cdot \ln \left(\sin ^{2}(y)+1\right)+c} \end{array} $$
