Question

若f(x)在[0,1]上连续,证明:

在线坐等 急

Answer 1

(1)证明：
令t=π/2-x, 则sinx=cost, dx=-dt
∫(0,π/2)f(sinx)dx
=-∫(π/2,0)f(cost)dt
=∫(0,π/2)f(cost)dt
=∫(0,π/2)f(cosx)dx
证毕

(2)证明：
令t=π-x, 则sinx=sint, dx=-dt
∫(0,π)xf(sinx)dx
=-∫(π,0)(π-t)f(sint)dt
=∫(0,π)(π-t)f(sint)dt
=π∫(0,π)f(sint)dt-∫(0,π)tf(sint)dt
=π∫(0,π)f(sinx)dx-∫(0,π)xf(sinx)dt
∴2∫(0,π)xf(sinx)dx=π∫(0,π)f(sinx)dx
∴∫(0,π)xf(sinx)dx=(π/2)∫(0,π)f(sinx)dx
证毕

∫(0,π) xsinx/(1+cos²x)dx
=(π/2)∫(0,π)sinx/(1+cos²x)dx
=-(π/2)arctan(cosx)|(0,π)
=-(π/2)[arctan(-1)-arctan1]
=-(π/2)[(-π/4)-(π/4)]
=π²/4

Answer 2