Question

高数证明

Answer 1

令t=x-π/2，dx=dt
∫(0,π) sin^nxdx=∫(-π/2,π/2) sin^n(t+π/2)dt
=∫(-π/2,π/2) cos^ntdt
=2*∫(0,π/2) cos^ntdt
∫(0,π) sin^nxdx-2*∫(0,π/2) sin^nxdx
=2*∫(0,π/2) cos^nxdx-2*∫(0,π/2) sin^nxdx
=2*∫(0,π/2) [cos^nx-sin^nx]dx
令x=π/2-t，dx=-dt
∫(0,π) sin^nxdx-2*∫(0,π/2) sin^nxdx
=2*∫(0,π/2) [cos^nx-sin^nx]dx
=2*∫(π/2,0) [cos^n(π/2-t)-sin^n(π/2-t)](-dt)
=2*∫(0,π/2) [sin^nt-cos^nt]dt
=-2*∫(0,π/2) [cos^nt-sin^nt]dt
所以2*∫(0,π/2) [sin^nx-cos^nx]dx=0
即∫(0,π) sin^nxdx-2*∫(0,π/2) sin^nxdx=0
∫(0,π) sin^nxdx=2*∫(0,π/2) sin^nxdx

Answer 2

这题还要证？将左边积分拆成[0,pi/2]+[pi/2,pi]两个积分，然后对后一积分作变量代换y=pi-x，再考率到sin(x)=sin(pi-x)，所以这后一积分就可化成前一积分。