∫(1/√2,1) √(1-x^2)/x^2 dx
换元法,x=sint
=∫(π/4,π/2) √(1-sin^2t)/sin^2t d(sint)
=∫(π/4,π/2) cos^2t / sin^2t dt
=∫(π/4,π/2) (1-sin^2t) / sin^2t dt
=∫(π/4,π/2) 1/sin^2t - 1 dt
=-cott-t | (π/4,π/2)
=(0-π/2)-(-1-π/4)
=1-π/4
有不懂欢迎追问
求定积分 (根号1-x^2)/x^2 (下限1/根号2 上限1)