Найти интеграл от (1/(cosx*sin^5x))dx

d(ctgx)=-dx/Sin²x

1/sin²x = 1 + ctg²x

$\int{\frac{1}{CosxSin^5x}}\, dx = \int{\frac{Sinx}{CosxSin^6x}}\, dx = - \int{\frac{1}{ctgxSin^4x}}\, d(ctgx)$

Делаем замену ctgx = t

$-\int{\frac{(1+t^2)^2}{t}}\, dt= -\int{\frac{(t^4 + 2t^2 + 1)}{t}}\, dt = -\int{(t^3 + 2t + \frac{1}{t})\, dt$

$-\int{(t^3 + 2t + \frac{1}{t})\, dt = -\frac{t^4}{4} - t^2 - ln|t| + C = - \frac{tg^4x}{4}-tg^2x-ln|tgx|+C$