Question

求解一道高数题，第三小题，谢谢

Answer 1

设x=1/t，则dx=-1/t²·dt
x=2时，t=1/2
x→+∞时，t→0

所以，
原式=∫(1/2~0)t²/√(1-t²)·(-1/t²)·dt
=∫(1/2~0)-1/√(1-t²)·dt
=∫(0~1/2)1/√(1-t²)·dt
=arcsint |(0~1/2)
=π/6-0
=π/6

追答

本题也可以应用三角换元，

设x=sect，则
dx=sect·tant·dt
x=2时，t=π/3
x→+∞时，t→π/2

所以，
原式=∫(π/3~π/2)1/(sect·tant)·sect·tant·dt
=∫(π/3~π/2)1dt
=t |(π/3~π/2)
=π/2-π/3
=π/6