Question

1-arcsinx/根号下1-x的平方的不定积分,大学微积分

Answer 1

∫(1-arcsinx)/√(1-x^2) dx

let

x=siny

dx =cosydy

∫(1-arcsinx)/√(1-x^2) dx

=∫(1-y) dy

=y - (1/2)y^2 + C

=arcsinx - (1/2)[arcsinx]^2 + C