f(x)=∫(0,x) cost / 1+sin^2t dt
則
∫(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^2t dt=0,積分上下限一樣,積分明顯為0
f(π/2)
=∫(0,π/2) cost / 1+sin^2t dt
=∫(0,π/2) 1 / 1+sin^2t d(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
有不懂歡迎追問