Calculus of a Single Variable, Chapter 8, 8.4, Section 8.4, Problem 23

Given ,
int sqrt(16-4x^2)dx
This Integral can be solved by using the Trigonometric substitutions (Trig substitutions)
For sqrt(a-bx^2) we have to take x= sqrt(a/b) sin(u)

so here , For
int sqrt(16-4x^2)dx -----(1)
x can be given as
x= sqrt(16/4) sin(u)= sqrt(4) sin(u) = 2sin(u)
so, x= 2sin(u) => dx = 2 cos(u) du
Now substituting x in (1) we get,
int sqrt(16-4x^2)dx
=int sqrt(16-4(2sin(u))^2) (2 cos(u) du)
= int sqrt(16-4*4(sin(u))^2) (2 cos(u) du)
= int sqrt(16-16(sin(u))^2) (2 cos(u) du)
= int sqrt(16(1-(sin(u))^2)) (2 cos(u) du)
= int sqrt(16(cos(u))^2) (2 cos(u) du)
= int (4cos(u)) (2 cos(u) du)
= int 8cos^2(u) du
= 8 int cos^2(u) du
= 8 int (1+cos(2u))/2 du
= (8/2) int (1+cos(2u)) du
= 4 int (1+cos(2u)) du
= 4 [int 1 du +int cos(2u) du]
= 4 [u+(1/2)(sin(2u))] +c
but x= 2sin(u)
=> (x/2)= sin(u)
=> u= sin^(-1) (x/2)
so,
4 [u+(1/2)(sin(2u))] +c
=4 [sin^(-1) (x/2)+1/2sin(2(sin^(-1) (x/2)))] +c
so,
int sqrt(16-4x^2)dx
=4sin^(-1) (x/2)+2sin(2(sin^(-1) (x/2))) +c