Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 12

int(du)/(usqrt(5-u^2))

Let
u=sqrt5sin(theta)
(du)/[d(theta)]=sqrt5cos(theta)
(du)=sqrt5cos(theta)d(theta)

int(du)/[usqrt(5-u^2)]
=int1/(sqrt5sin(theta))*[sqrt(5)cos(theta)d(theta)]/sqrt[5-(sqrt5sin(theta)^2)]
=int[cot(theta)d(theta)]/sqrt(5-5sin^2theta)
=int[cot(theta)d(theta)]/sqrt[5(1-sin^2theta)]
=int[cot(theta)d(theta)]/sqrt(5cos^2theta)
=int[cot(theta)d(theta)]/[sqrt(5)cos(theta)]
=int[cos(theta)d(theta)]/[sqrt(5)sin(theta)cos(theta)]
=int[d(theta)]/[sqrt(5)sin(theta)]
=int[csc(theta)d(theta)]/sqrt(5)
=1/sqrt(5)intcsc(theta)d(theta)
=1/sqrt(5)ln|sqrt(5)/u-sqrt(5-u^2)/u|+C
=1/sqrt(5)*ln|[sqrt5-sqrt(5-u^2)]/u|+C

The final answer is
=1/sqrt(5)*ln|[sqrt5-sqrt(5-u^2)]/u|+C