Question

用换元法求不定积分∫sin2x×3^cosx^2×二次根号（1+3^cosx^2×）dx..

Answer 1

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