We did this exercise in class in a way, but at home I tried to solve it in a different way and I do not know if it is right or wrong. May you help me please?
$\mathbf{\int tan^{5}x \, \, \, sec^{4}x \; \; dx} \\ $
$\mathbf{\int tan^{4}x \, \, \, sec^{3}x \; \; tanx \, secx \, \, \, dx} \\ $
$\mathbf{\int (sec^{2}x - 1)^{2} \, \, \, sec^{3}x \; \; tanx \, secx \, \, \, dx} \\$
$ \qquad u = secx \\ \\ \qquad du = secx \, tanx \, dx \\$
$\mathbf{\int (u^{2} - 1)^{2} \, \, \, u^{3} \; \; du} \\$
$\mathbf{\int (u^{4} - 2u^{2} + 1) \, \, \, u^{3} \; \; du} \\$
$\mathbf{\int (u^{7} - 2u^{5} + u^{3}) \, \, du} \\$
$\mathbf{\frac{u^{8}}{8}-2\frac{u^{6}}{6}+\frac{u^{4}}{4}} + C \\$
$\mathbf{\frac{sec^{8}x}{8}-1\frac{sec^{6}x}{3}+\frac{sec^{4}x}{4}} + C \\$
Thanks a lot.