Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 35

Use the arc length formula,
L=intds
ds=sqrt(1+(dy/dx)^2)dx , if y=f(x), a<= x<= b
Given x^2+y^2=9
=>y^2=9-x^2
y=(9-x^2)^(1/2)
dy/dx=1/2(9-x^2)^(1/2-1)*(-2x)
=-x/sqrt(9-x^2)
Plug in the above in ds,
ds=sqrt(1+(-x/sqrt(9-x^2)))^2dx
ds=sqrt(1+x^2/(9-x^2))dx
ds=sqrt((9-x^2+x^2)/(9-x^2))dx
ds=3/sqrt(9-x^2)dx
The limits are x=0 and x=2,
L=int_0^2 3/sqrt(9-x^2)dx
=3int_0^2 1/sqrt(9-x^2)dx
Now let's first evaluate the indefinite integral by using integral substitution,
Let x=3sin(u)
dx=3cos(u)du
int1/sqrt(9-x^2)dx=int1/sqrt(9-(3sin(u))^2)3cos(u)du
=int(3cos(u))/sqrt(9-9sin^2(u))du
=int(3cos(u))/(sqrt(9)sqrt(1-sin^2(u)))du
=int(3cos(u))/(3cos(u))du
=int1du
=u
substitute back u and add a constant C to the solution,
=arcsin(x/3)+C
L=3[arcsin(x/3)]_0^2
L=3[arcsin(2/3)-arcsin(0)]
L=3arcsin(2/3)
~~2.19