Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 10

intsin^-1xdx
If f(x) and g(x) are differentiable functions, then
intf(x)g'(x)dx=f(x)g(x)-intf'(x)g(x)dx
If we write f(x)=u and g'(x)=v, then
intuvdx=uintvdx-int(u'intvdx)dx
Using the above method of integration by parts,
intsin^-1xdx=sin^-1x*int1dx-int(d/dx(sin^-1x)int1dx)dx
=sin^-1x*x-int(1/sqrt(1-x^2)*x)dx
=xsin^-1x-intx/sqrt(1-x^2)dx
Now evaluate using the method of substitution,
Substitute t=1-x^2,=> dt=-2xdx
intx/sqrt(1-x^2)dx=intdt/(-2sqrt(t))
=-1/2intdt/sqrt(t)
=-1/2(t^(-1/2+1)/(-1/2+1))
=-1/2(t^(1/2)/(1/2))
=-t^(1/2)
substitute back t=1-x^2
=-(1-x^2)^(1/2)
intsin^-1xdx=xsin^-1x-(-(1-x^2)^(1/2))
adding constant C to the solution,
=xsin^-1x+sqrt(1-x^2)+C