Given: #int sqrt((9-w^2)/w^2)dw#
#int sqrt((9-w^2)/w^2)dw = int sqrt(9-w^2)/wdw#
Let #w = 3sin(x)#, then #dw = 3cos(x)dx#
#int sqrt((9-w^2)/w^2)dw = int sqrt(9-9sin^2(x))/(3sin(x))3cos(x)dx#
Bring the 9 outside of the radical as 3:
#int sqrt((9-w^2)/w^2)dw = int (3sqrt(1-sin^2(x)))/(3sin(x))3cos(x)dx#
Move the remaining 3 outside of the integral:
#int sqrt((9-w^2)/w^2)dw = 3int (sqrt(1-sin^2(x)))/sin(x)cos(x)dx#
Substitute #cos(x) = sqrt(1-sin^2(x))#:
#int sqrt((9-w^2)/w^2)dw = 3int cos^2(x)/sin(x)dx#
Substitute #cos^2(x) = 1-sin^2(x)#:
#int sqrt((9-w^2)/w^2)dw = 3int (1-sin^2(x))/sin(x)dx#
Perform the division:
#int sqrt((9-w^2)/w^2)dw = 3int 1/sin(x)-sin(x)dx#
Substitute #1/sin(x) = csc(x)# and separate into two integrals:
#int sqrt((9-w^2)/w^2)dw = 3int csc(x)-3int sin(x)dx#
We know these two integrals:
#int sqrt((9-w^2)/w^2)dw = 3ln( cot(x)+csc(x))+3cos(x)+C#
Substitute #x = sin^-1(w/3)#
#int sqrt((9-w^2)/w^2)dw = 3ln( cot(sin^-1(w/3))+csc(sin^-1(w/3)))+3cos(sin^-1(w/3))+C#
Substitute #cot(sin^-1(w/3))= sqrt(9-w^2)/w#
#int sqrt((9-w^2)/w^2)dw = 3ln( sqrt(9-w^2)/w+csc(sin^-1(w/3)))+3cos(sin^-1(w/3))+C#
Substitute #csc(sin^-1(w/3))= 3/w#
#int sqrt((9-w^2)/w^2)dw = 3ln( sqrt(9-w^2)/w+3/w)+3cos(sin^-1(w/3))+C#
Substitute #cos(sin^-1(w/3)) = sqrt(1-w^2/9)#
#int sqrt((9-w^2)/w^2)dw = 3ln( sqrt(9-w^2)/w+3/w)+3sqrt(1-w^2/9)+C#
Multiply the last term through:
#int sqrt((9-w^2)/w^2)dw = 3ln( sqrt(9-w^2)/w+3/w)+sqrt(9-w^2)+C#
