請教積分題目

Let x=sinu, then
arcsinx = u
dx = cos u du
1-x^2 = cos^2 u
x=0, u=0
x=1/√2, u=π/4.

So ∫(0 to 1/√2) arcsinx dx /(1-x^2)^(3/2)
=∫(0 to π/4) u cos u du / cos^3 u
=∫(0 to π/4) u sec^2 u du
=∫(0 to π/4) u d(tan u)
=(π/4)tan(π/4) - (0)tan(0) - ∫(0 to π/4) tan u du
=π/4 + ln|cos π/4| - ln|cos0|
=π/4 + ln(1/√2) - ln0
=π/4 - (1/2)ln2