How do you integrate #int sqrt((9-w^2)/w^2)dw#?

2 Answers
Feb 7, 2018

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#

Feb 7, 2018

The answer is #=sqrt(9-w^2)+3/2ln(|sqrt(9-w^2)-3|)-3/2ln(sqrt(9-w^2)+3)+C#

Explanation:

Perform the substitution

#u=w^2#, #=>#, #du=2wdw#, #dw=1/(2w)du#

#I=intsqrt((9-w^2)/(w^2))dw=intsqrt((9-w^2))/(w)dw#

#=intsqrt(9-u)/w*1/(2w)du#

#=1/2intsqrt(9-u)/udu#

Let

#v=sqrt(9-u)#, #=>#, #dv=-1/(2sqrt(9-u))du#

#I=1/2int(v*2v)/(v^2-9)dv#

#=intv^2/(v^2-9)dv#

#=int(1+9/(v^2-9))dv#

Perform the decomposition into partial fractions

#9/(v^2-9)=9/((v+3)(v-3))=A/(v+3)+B/(v-3)#

#9=A(v-3)+B(v+3)#

Let #v=-3#, #9=-6A#, #A=-3/2#

Let #v=3#, #9=6B#, #B=3/2#

Therefore,

#I=int(1+3/(2(v-3))-3/(2(v+3)))dv#

#=v+3/2ln(v-3)-3/2ln(v+3)#

#=sqrt(9-u)+3/2ln(sqrt(9-u)-3)-3/2ln(sqrt(9-u)+3)#

#=sqrt(9-w^2)+3/2ln(|sqrt(9-w^2)-3|)-3/2ln(sqrt(9-w^2)+3)+C#