Integral (sin3x+cos3x) ²dx

А) Int (3/[sin (3x) ]^2-cos (2x)) dx = Int 3/[sin (3x) ]^2 dx - Int cos (2x) dx =

Int d (3x) / [sin (3x) ]^2 - 0.5*Int cos (2x) d (2x) = - ctg (3x) - 0.5*sin (2x) + C.

-ctg (-3*pi/2) - 0.5*sin (-2*pi/2) + C=3

0 0 C=3

Answer: - ctg (3x) - 0.5*sin (2x) + 3

Б) 2Pin, n прин. Z x = + - (Pi - arccos (1/3)) + 2Pin, n прин. Z 3) sqrt (3) * cosx = sinx |:cosx tgx = sqrt (3) x = arctg (sqrt (3)) + Pik, k прин. Z x = Pi/3 + Pik, k прин

Точно не уверенa!