`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`