Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 31

int_0^(1/2)cos^(-1)xdx
Let's first evaluate the indefinite integral by using the method of integration by parts,
intcos^(-1)xdx=cos^(-1)x*int1dx-int(d/dx(cos^(-1)x)int1dx)dx
=cos^(-1)x*x-int(-1/sqrt(1-x^2)*x)dx
=xcos^(-1)x+intx/sqrt(1-x^2)dx
Now let's evaluate intx/sqrt(1-x^2)dx using the method of substitution,
Let's 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
=-sqrt(1-x^2)
:.intcos^(-1)xdx=xcos^(-1)x-sqrt(1-x^2)+C
Now let's evaluate the definite integral,
int_0^(1/2)cos^(-1)x=[xcos^(-1)x-sqrt(1-x^2)]_0^(1/2)
=[1/2cos^(-1)1/2-sqrt(1-(1/2)^2)]-[0cos^(-1)0-sqrt(1-0^2)]
=[1/2*pi/3-sqrt(1-1/4)]-[-1]
=pi/6-sqrt(3)/2+1