Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 6

int[(sin^3sqrtx)/(sqrtx)]dx=
Integrate using the u-substitution method. For this problem the u-substitution method will be used twice. The first time we substitute, let's use the variable y.
Let
y=sqrtx
(dy)/dx=1/(2sqrtx)
dx=2sqrtxdy

int[(sin^3(y))/y*2sqrtxdy=
2int[(sin^3(y))/(y)]*ydy=
2intsin^3(y)dy=
2intsin^2(y)sin(y)dy=
2int(1-cos^2(y))sin(y)dy=
The u-substitution method will be used a second time. We will use the variable u.
Let
u=cosy
(du)/dy=-sin(y)
dy=(-sin(y))/(du)

2int(1-u^2)sin(y)[(du)/(-sin(y))]=
-2int(1-u^2)du=
-2[u-1/3u^3]+C=
-2u+2/3u^3+C

Substitute in for u. u=cos(y)
-2cos(y)+2/3cos^3(y)+C

Substitute in for y. y=sqrtx
-2cos(sqrtx)+2/3cos^3(sqrtx)+C

The final answer is: -2cos(sqrtx)+2/3cos^3(sqrtx)+C