证明：
因为∫(0→π)f(sinx)dx=∫(0→π/2)f(sinx)dx+∫(π/2→π)f(sinx)dx
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(sint)dt
