三角函数的定积分问题∫（上限为2,下限为-2）√（4-x^2)再乘以（sinx+1)dx,该题根号里面的式子能不能用令X

可以,不过先用奇偶性不是更快一点吗?
∫(- 2→2) √(4 - x²)(sinx + 1) dx
= ∫(- 2→2) √(4 - x²)sinx dx + ∫(- 2→2) √(4 - x²) dx,前奇后偶
= 0 + 2∫(0→2) √(4 - x²) dx
= 2∫(0→2) √(4 - x²) dx
令x = 2sint,dx = 2cost dt
x = 0 ==> t = 0
x = 2 ==> sint = 1 ==> t = π/2
= 2∫(0→π/2) √(4 - 4sin²t)(2cost dt)
= 2∫(0→π/2) (2cost)² dt
= 8∫(0→π/2) (1 + cos2t)/2 dt
= 4∫(0→π/2) (1 + cos2t) dt
= 4[t + (1/2)sin2t] |(0→π/2)
= 4(π/2 + 0)
= 2π