Помогите решить 2 и 8 интеграл.

2.
$\int\limits { \frac{ \sqrt[6]{ctgx} }{cos^2 x} } \, dx = \int\limits { \frac{ (ctgx)^{ \frac{1}{6} } }{cos^2 x} } \, dx = - \int\limits { \frac{ t^{ \frac{1}{6} }sin^2x }{cos^2 x} } \, dt = - \int\limits { \frac{ t^{ \frac{1}{6} }}{ \frac{cos^2 x}{sin^2x} } } \, dt = \\ \\ t=ctgx \\ dt=- \frac{dx}{sin^2 x } \\ dx=-dt*sin^2 x \\ \\ = - \int\limits { \frac{ t^{ \frac{1}{6} }}{ t^2 } } \, dt = - \int\limits { t^{- \frac{11}{6} } } \, dt = - \frac{1}{- \frac{11}{6} +1} t^{- \frac{11}{6} +1} +C =$

$= - \frac{1}{- \frac{5}{6}} t^{- \frac{5}{6}} +C = \frac{6}{5} \frac{1}{t^{ \frac{5}{6} }}+ C = \frac{6}{5} \frac{1}{ctg^{ \frac{5}{6} }x} +C$

8.
$\int\limits { \frac{1}{sin^2 5x cos^ 5x} } \, dx = 4 \int\limits { \frac{1}{sin^2 10x} } \, dx = \frac{4}{10} \int\limits { \frac{1}{sin^2 10x} } \, d(10x) = - \frac{2}{5} ctg10x +C$