Question

Найти интеграл от (1/(cosx*sin^5x))dx

Answer 1

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

 

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

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