Question

求积分详细步骤

Answer 1

令x=√2sina
dx=√2cosada
√(2-x²)=√2cosa

x=1,a=π/6
x=0,a=0
原式=2∫(0到π/6)2sin²a*√2cosa*√2cosada
=2∫(0到π/6)(4sin²acos²a)da
=2∫(0到π/6)sin²2ada
=∫(0到π/6)(1-cos4a)/2*(1/2 d4a)
=1/4*(4a-sin4a)(0到π/6)
=π/6-√3/8

追问

答案是圆周率/4

追答

哦，对了
x=1时，sina=1/√2
则a=π/4，不是π/6

令x=√2sina
dx=√2cosada
√(2-x²)=√2cosa

x=1,a=π/4
x=0,a=0
原式=2∫(0到π/4)2sin²a*√2cosa*√2cosada
=2∫(0到π/4)(4sin²acos²a)da
=2∫(0到π/4)sin²2ada
=∫(0到π/4)(1-cos4a)/2*(1/2 d4a)
=1/4*(4a-sin4a)(0到π/4)
=1/4*(π-sinπ)-1/4*(0-0)
=π/4