Calculus of a Single Variable, Chapter 8, 8.1, Section 8.1, Problem 43

int6/sqrt(10x-x^2)dx
Take the constant out ,
=6int1/sqrt(10x-x^2)dx
Let's rewrite the denominator by completing the square,
=6int1/(sqrt(-(x-5)^2+25))dx
Now apply integral substitution: u=x-5
du=1dx
=6int1/sqrt(-u^2+25)dx
=6int1/sqrt(5^2-u^2)du
Again apply integral substitution: u=5sin(v)
=>du=5cos(v)dv
=6int(1/sqrt(5^2-5^2sin^2(v)))5cos(v)dv
=6int(5cos(v))/(sqrt(5^2)sqrt(1-sin^2(v)))dv
Use the trigonometric identity: cos^2(x)=1-sin^2(x)
=6int(5cos(v))/(5sqrt(cos^2(v)))dv
=6int(5cos(v))/(5cos(v))dv , assuming cos(v)>=0
=6int1dv
=6v
substitute back v=arcsin(u/5) and u=(x-5)
=6arcsin((x-5)/5)
Add a constant C to the solution,
=6arcsin((x-5)/5)+C