Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 19

given to solve ,
int sqrt(25-x^2) dx
using the Trig Substitutions we can solve these type of integrals easily and the solution is as follows

for sqrt(a-bx^2) we can take x= sqrt(a/b) sin(u)
so ,
int sqrt(25-x^2) dx
the x = sqrt(25/1) sin(u) = 5sin(u)
=> dx = 5 cos(u) du
so ,
int sqrt(25-x^2) dx
= int sqrt(25-(5sin(u))^2) (5 cos(u) du)
= int sqrt(25-25(sin(u))^2) (5 cos(u) du)
= int 5 sqrt(1-sin^2 u )(5 cos(u) du)
= int 25 (cos(u))(cos(u)) du
= 25 int cos^2(u) du
= 25 int(1+cos(2u))/2 du
= (25/2) int (1+cos(2u)) du
= (25/2) [u+(1/2)(sin(2u))]+c
but x=5sin(u)
=> x/5 = sin(u)
=> u= arcsin(x/5)
so,
(25/2) [u+(1/2)(sin(2u))]+c
=(25/2) [(arcsin(x/5))+(1/2)(sin(2(arcsin(x/5))))]+c