Здравствуйте, Попов Антон Андреевич.

Интеграл имеет вид:

∫ dx / (2sin(x) + cos(x) + 2)

1. Здесь надо применить тригонометрическую подстановку:

tg(x/2) = t

Тогда:

x = 2*arctg(t)

[$8658$] dx = [2*arctg(t)]'dt = [2*dt / (1 + t²)]

Также:

cos(x) = [cos(2*(x/2))] / 1 = [cos²(x/2) - sin²(x/2)] / [cos²(x/2) + sin²(x/2)] =

= {[cos²(x/2) - sin²(x/2)] : cos²(x/2)} / {[cos²(x/2) + sin²(x/2)] : cos²(x/2)} =

= [1 - tg²(x/2)] / [1 + tg²(x/2)] = [1 - t²] / [1 + t²]

sin(x) = [sin(2*(x/2))] / 1 = [2*cos(x/2)*sin(x/2)] / [cos²(x/2) + sin²(x/2)] =

= {[2*cos(x/2)*sin(x/2)] : cos²(x/2)} / {[cos²(x/2) + sin²(x/2)] : cos²(x/2)} =

= [2*tg(x/2)] / [1 + tg²(x/2)] = [2t] / [1 + t²]

Итак:

∫ dx / (2sin(x) + cos(x) + 2) = ∫ {2*dt / (1 + t²)} / {[2*(2t) / (1 + t²)] + [(1 - t²) / (1 + t²)] + 2} =

= ∫ {2*dt} / {t² + 4t + 3}

2. Далее раскладываем полученную дробь

Так как:

t² + 4t + 3 = 0

при t_1,2 = - 2 [$177$] [$8730$]((- 2)² - 3) = - 2 [$177$] 1

то:

t² + 4t + 3 = (t + 1)*(t + 3)

Пусть:

2 / (t² + 4t + 3) = {A / (t + 1)} + {B / (t + 3)}

Тогда:

2 = A*(t + 3) + B*(t + 1) = (A + B)*t + (3A + B)

Получим систему уравнений:

{ A + B = 0
{ 3A + B = 2

Решая, получим: A = 1 и B = - 1

Значит:

2 / (t² + 4t + 3) = {1 / (t + 1)} - {1 / (t + 3)}

3. Продолжаем решение интеграла

∫ dx / (2sin(x) + cos(x) + 2) = ∫ {2*dt} / {t² + 4t + 3} = ∫ {dt / (t + 1)} - ∫ {dt / (t + 3)} =

= ln |t + 1| - ln |t + 3| + ln (C) = ln |C * (t + 1) / (t + 3)| =

= / так как t = tg(x/2), то / =

= ln |C * (tg(x/2) + 1) / (tg(x/2) + 3)| = ln|C * [(tg(x/2) + 1)*cos(x/2)] / [(tg(x/2) + 3)*cos(x/2)]| =

= ln |C * [sin(x/2) + cos(x/2)] / [sin(x/2) + 3*cos(x/2)]|, где C = const

Ответ:

∫ dx / (2sin(x) + cos(x) + 2) = ln |C * [sin(x/2) + cos(x/2)] / [sin(x/2) + 3*cos(x/2)]|, где C = const