Question

数学求定积分

Answer 1

∫(0,π/2) cos^nxdx
=∫(0,π/2) cos^(n-1)x*cosxdx
=∫(0,π/2) cos^(n-1)xd(sinx)
=cos^(n-1)xsinx|(0,π/2)+∫(0,π/2) (n-1)sin^2xcos^(n-2)xdx
=(n-1)*∫(0,π/2) (1-cos^2x)cos^(n-2)xdx
=(n-1)*∫(0,π/2) cos^(n-2)xdx-(n-1)*∫(0,π/2) cos^nxdx
所以∫(0,π/2) cos^nxdx=(n-1)/n*∫(0,π/2) cos^(n-2)xdx
原式=∫(0,π/2) cos^8xdx
=7/8*5/6*3/4*∫(0,π/2) cos^2xdx
=35/64*1/2*∫(0,π/2) (1+cos2x) dx
=35/128*(x+1/2*sin2x)|(0,π/2)
=35/128*(π/2)
=35π/256

追问

差了一点……但是有思路我就懂啦^_^

追答

额~~你给出的题目的积分上限是π/2，可不是π哦~~

追问

啊啊啊！！！我错啦=_=😁

追答

嘿嘿