Question

高等数学，图中第十题

Answer 1

约定:∫[a,b]表示[a,b]上的定积分
原式左边=∫[0,π](sinx)^ndx
=∫[-π/2,π/2](sin(t+(π/2))^nd(t+(π/2)) 设x=t+(π/2)
=∫[-π/2,π/2](cost)^ndt
=2∫[0,π/2](cost)^ndt (y=(cost)^n是偶函数)
=2∫[π/2,0](cos((π/2)-u))^nd(π/2)-u) 设t=(π/2)-u
=-2∫[π/2,0](sinu)^ndu
=2∫[0,π/2](sinu)^ndu
=2∫[0,π/2](sinx)^ndx
=原式右边
所以∫[0,π](sinx)^ndx=2∫[0,π/2](sinx)^ndx

希望能帮到你！