Question

高数题，求解

给个过程，谢谢

Answer 1

约定:∫[a,b] 表示求[a,b]区间上的定积分
因∫[π/2,π]xf(sinx)dx (令x=π-t)
=∫[π/2,0](π-t)f(sin(π-t))d(π-t)
=-∫[π/2,0](π-t)f(sint)dt
=∫[0,π/2](π-t)f(sint)dt
=π∫[0,π/2]tf(sint)dt-∫[0,π/2]tf(sint)dt
即∫[π/2,π]xf(sinx)dx=π∫[0,π/2]xf(sinx)dx-∫[0,π/2]xf(sinx)dx
得∫[0,π]xf(sinx)dx=∫[0,π/2]xf(sinx)dx+∫[π/2,π]xf(sinx)dx
=∫[0,π/2]xf(sinx)dx+π∫[0,π/2]xf(sinx)dx-∫[0,π/2]xf(sinx)dx
=π∫[0,π/2]xf(sinx)dx
所以∫[0,π]xf(sinx)dx=π∫[0,π/2]xf(sinx)dx

Answer 2

用t=pi-x带入得到
原积分A=∫(pi-t)f(sin(pi-t))d(pi-t) |t=pi,0 = -∫pif(sint) dt|t=pi,0 +∫tf(sint)dt|t=pi,0
=pi∫f(sint)dt |t=0,pi -∫tf(sint)dt|t=0,pi
=pi∫f(sint)dt |t=0,pi -A
所以A=0.5 pi∫f(sint)dt |t=0,pi