Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 7

Given to solve,
int sin^3 (2 theta) sqrt(cos(2 theta)) d(theta)
let x= theta (just for convinence)
so,
int sin^3 (2 theta) sqrt(cos(2 theta)) d(theta)
=int sin^3 (2x) sqrt(cos(2x)) dx

let 2x= u so , du = 2dx then ,
int sin^3 (2x) sqrt(cos(2x)) dx
=int sin^3 (u) sqrt(cos(u)) (du)/2
=(1/2)int sin^2 (u) sin u sqrt(cos(u)) du
= (1/2)int (1-cos^2 (u)) sin u sqrt(cos(u)) du

let cos u =t, so , dt = -sin(u) du
then,
(1/2)int (1-cos^2 (u)) sin u sqrt(cos(u)) du
= (1/2)int (1-t^2) sqrt(t) sin u du
=(1/2)int (1-t^2) sqrt(t) (-dt)
= (-1/2)int (1-t^2) sqrt(t) (dt)
= (-1/2) int (t^(1/2) - t^(5/2))dt
= (-1/2) [(t^(3/2))/(3/2) - t^((5/2)+1)/((5/2)+1)]
= (-1/2) [(t^(3/2))/(3/2) - (t^(7/2))/(7/2)]
but t= cos u = cos(2x) so,
= (-1/2) [((cos(2x))^(3/2))/(3/2) - ((cos(2x))^(7/2))/(7/2)]
= (1/2)[((cos(2x))^(7/2))/(7/2) -((cos(2x))^(3/2))/(3/2)]
but x= theta, so
= (1/2)[((cos(2(theta)))^(7/2))/(7/2) -((cos(2(theta)))^(3/2))/(3/2)]
so,
int sin^3 (2 theta) sqrt(cos(2 theta)) d(theta)
=(1/2)[((cos(2(theta)))^(7/2))/(7/2) -((cos(2(theta)))^(3/2))/(3/2)]