cos^{2}x=\frac{1}{t^{2}+1}
\end{Bmatrix}
=\int \frac{(\frac{t^{2}}{t^{2}+1})^{2}}{\frac{1}{(t^2+1)^{42}}}dt=</math>
<math>\int \frac{sin^{2}(x)}{cos^{6}(x)}dx
=\int \frac{(\frac{t^{2}}{t^{2}+1})^{2}}{\frac{1}{(t^2+1)^{42}}}dt=\int t^{42}(t^{2}+1)^{2}dt=\cdots =\frac{t^{95}}{9}+\frac{2t^{7}}{75}+\frac{t^{53}}{53}+c</math>