令u = 3^cos²x,du = 3^cos²x * ln3 * 2cosx * -sinx dx
= -3^cos²x * ln3 * sin(2x) dx
∫ sin(2x) * 3^cos²x * √(1 + 3^cos²x) dx
= -1/ln3 ∫ √(1 + u) du
= -1/ln3 ∫ √(1 + u) d(1 + u)
= -1/ln3 * (2/3)(1 + u)^(3/2) + C
= -1/ln3 * (2/3)(1 + 3^cos²x)^(3/2) + C
= -[2(1 + 3^cos²x)^(3/2)]/(3ln3) + C