How do you integrate #int 1/(x-sqrt(9 + x^2))dx# using trigonometric substitution?
1 Answer
Explanation:
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=intsqrt(9+9tan^2theta)(3sec^2thetad theta)#
#=9intsqrt(1+tan^2theta)(sec^2thetad theta)#
Since
#J=9intsec^3thetad theta#
To solve this integral, we will use integration by parts, which takes the form
#{(u=sectheta" "=>" "du=secthetatanthetad theta),(dv=sec^2thetad theta" "=>" "v=tantheta):}#
Thus:
#K=secthetatantheta-intsecthetatan^2thetad theta#
Let
#K=secthetatantheta-intsectheta(sec^2theta-1)d theta#
#K=secthetatantheta-intsec^3thetad theta+intsecthetad theta#
Notice that we have the original integral
#K=secthetatantheta-K+lnabs(sectheta+tantheta)#
Solving for
#2K=secthetatantheta+lnabs(sectheta+tantheta)#
#K=(secthetatantheta+lnabs(sectheta+tantheta))/2=intsec^3thetad theta#
Returning to
#J=9/2(secthetatantheta+lnabs(sectheta+tantheta))#
We should write this in terms of just
#J=9/2(sqrt(tan^2theta+1)(tantheta)+lnabs(sqrt(tan^2theta+1)+tantheta))#
Since
#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
#intsqrt(9+x^2)dx=x/2sqrt(x^2+9)+9/2lnabs(sqrt(x^2+9)+x)#
Plugging this into
#I=-1/18x^2-1/9(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#