Question

定积分。怎么来的？能给详细过程吗？

Answer 1

1.证明∫(0→π)xf(sinx)dx=(π/2)∫(0→π)f(sinx)dx
令x=π-u，dx=-du
当x=π时，u=0
当x=0时，u=π
∫(0→π)xf(sinx)dx
=∫(π→0)(π-u)f[sin(π-u)]*(-du)
=∫(0→π)(π-u)f[sin(π-u)]du
=∫(0→π)πf[sin(π-u)]du-∫(0→π)uf[sin(π-u)]du
=π∫(0→π)f(sinu)du-∫(0→π)uf(sinu)du
(定积分与积分变量字母无关）
=π∫(0→π)f(sinx)dx-∫(0→π)xf(sinx)dx
移项
2∫(0→π)xf(sinx)dx=π∫(0→π)f(sinx)dx
所以∫(0→π)xf(sinx)dx=(π/2)∫(0→π)f(sinx)dx

2.证明(π/2)∫(0→π)f(sinx)dx=π∫(0→π/2)f(sinx)dx
实际上就是证明∫(0→π)f(sinx)dx=2∫(0→π/2)f(sinx)dx
因为∫(0→π)f(sinx)dx=∫(0→π/2)f(sinx)dx+∫(π/2→π)f(sinx)dx
只计算右半部分
令x=π-t 则当x=π/2时 t=π/2 当x=π时 t=0
所以∫(π/2→π)f(sinx)dx
=∫(π/2→0)f(sin(π-t))d(π-t)
=-∫(π/2→0)f(sint)dt
=∫(0→π/2)f(sint)dt
=∫(0→π/2)f(sinx)dx(定积分与积分变量字母无关）
于是∫(0→π)f(sinx)dx
=∫(0→π/2)f(sinx)dx+∫(π/2→π)f(sinx)dx
=2∫(0→π/2)f(sinx)dx

3.证明π∫(0→π/2)f(sinx)dx=π∫(0→π/2)f(cosx)dx
实际上就是证明∫(0→π/2)f(sinx)dx=∫(0→π/2)f(cosx)dx
令x=π/2-y，dx=-dy
当x=π/2时，y=0
当x=0时，y=π/2
∫(0→π/2)f(sinx)dx
=∫(π/2→0)f[sin(π/2-y)]*(-dy)
=∫(0→π/2)f(cosy)dy
(定积分与积分变量字母无关）
=∫(0→π/2)f(cosx)dx