f(x)=∫(0,x)cost/1+sin^2tdt
则
∫(0,π/2)f'(x)/1+f^2(x)dx
=∫(0,π/2)1/1+f^2(x)d(f(x))
=arctan(f(x))|(0,π/2)
=arctan(f(π/2))-arctan(f(0))
又有:
f(0)=∫(0,0)cost/1+sin^2tdt=0,积分上下限一样,积分明显为0
f(π/2)
=∫(0,π/2)cost/1+sin^2tdt
=∫(0,π/2)1/1+sin^2td(sint)
=arctan(sint)|(0,π/2)
=arctan(1)-arctan(0)
=π/4-0
=π/4
因此,
∫(0,π/2)f'(x)/1+f^2(x)dx
=arctan(f(π/2))-arctan(f(0))
=π/4-0
=π/4
有不懂欢迎追问