Question

请问这两道题怎么解？

Answer 1

(6)
令x=sint
t=arcsinx
当x=1时，t=π/2
当x=1/√2时，t=π/4
上下限变为(π/2，π/4)
dx=costdt
原式=∫cost*costdt/sin²t
=∫cos²tdt/sin²t
=∫(1-sin²t)dt/sin²t
=∫csc²tdt-∫dt
=-cott-t+C |(π/2，π/4)
=0-π/2-(-1-π/4)
=1-π/4
(8)
设u=x³
du=3x²dx
x²dx=du/3
∫x²dx/(1+x^6)
=∫du/3(1+u²)
=(1/3)*arctanu+C
=(1/3)*arctanx³+C |(1,0)
=(1/3)*(π/4-0)
=π/12