How do you integrate #int 1/(x-sqrt(9 + x^2))dx# using trigonometric substitution?

Question

Ezekiel Alexander · Accepted Answer

#-(x^2+xsqrt(x^2+9)+9lnabs(sqrt(x^2+9)+x))/18+C#

First multiply by the conjugate:

#I=int1/(x-sqrt(9+x^2))*(x+sqrt(9+x^2))/(x+sqrt(9+x^2))dx#

#=int(x+sqrt(9+x^2))/(x^2-(9+x^2))dx#

#=-1/9int(x+sqrt(9+x^2))dx#

#=-1/9intxdx-1/9intsqrt(9+x^2)dx#

#=-1/18x^2-1/9intsqrt(9+x^2)dx#

Letting #J=intsqrt(9+x^2)dx#, we will solve this by using the substitution #x=3tantheta#. This implies that #dx=3sec^2thetad theta#.

#J=intsqrt(9+x^2)dx=intsqrt(9+9tan^2theta)(3sec^2thetad theta)#

#=9intsqrt(1+tan^2theta)(sec^2thetad theta)#

Since #1+tan^2theta=sec^2theta#:

#J=9intsec^3thetad theta#

To solve this integral, we will use integration by parts, which takes the form #intudv=uv-intvdu#. For #K=intsec^3thetad theta#, let:

#{(u=sectheta" "=>" "du=secthetatanthetad theta),(dv=sec^2thetad theta" "=>" "v=tantheta):}#

Thus:

#K=secthetatantheta-intsecthetatan^2thetad theta#

Let #tan^2theta=sec^2theta-1#:

#K=secthetatantheta-intsectheta(sec^2theta-1)d theta#

#K=secthetatantheta-intsec^3thetad theta+intsecthetad theta#

Notice that we have the original integral #K=intsec^3thetad theta# imbedded in this integral. Also, the integration of just secant is a common integral.

#K=secthetatantheta-K+lnabs(sectheta+tantheta)#

Solving for #K#:

#2K=secthetatantheta+lnabs(sectheta+tantheta)#

#K=(secthetatantheta+lnabs(sectheta+tantheta))/2=intsec^3thetad theta#

Returning to #J=9intsec^3thetad theta#, we see that:

#J=9/2(secthetatantheta+lnabs(sectheta+tantheta))#

We should write this in terms of just #tantheta#:

#J=9/2(sqrt(tan^2theta+1)(tantheta)+lnabs(sqrt(tan^2theta+1)+tantheta))#

Since #tantheta=x/3#:

#J=9/2(sqrt(x^2/9+1)(x/3)+lnabs(sqrt(x^2/9+1)+x/3))#

#J=9/2(sqrt(1/9(x^2+9))(x/3)+lnabs(sqrt(1/9(x^2+9))+x/3))#

#J=9/2(x/9sqrt(x^2+9)+lnabs(1/3(sqrt(x^2+9)+x)))#

#J=x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x)-9/2ln(3)#

(Ignore the constant since this is from an integral, just we haven't yet added #C# for convenience purposes.)

#intsqrt(9+x^2)dx=x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x)#

Plugging this into #I=-1/18x^2-1/9intsqrt(9+x^2)dx#:

#I=-1/18x^2-1/9(x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x))#

#I=-1/18x^2-1/18xsqrt(x^2+9)-1/2lnabs(sqrt(x^2+9)+x)#

#I=-(x^2+xsqrt(x^2+9)+9lnabs(sqrt(x^2+9)+x))/18+C#