How do you integrate #int 1/sqrt(x^2-16x+3) # using trigonometric substitution?

2 Answers
Eric S.
Mar 14, 2018

Use the substitution #x-8=sqrt61sectheta#.

Explanation:

Let

#I=int1/sqrt(x^2-16x+3)dx#

Complete the square in the square root:

#I=int1/sqrt((x-8)^2-61)dx#

Apply the substitution #x-8=sqrt61sectheta#:

#I=int1/(sqrt61tantheta)(sqrt61secthetatanthetad theta)#

Simplify:

#I=intsecthetad theta#

Integrate directly:

#I=ln|sectheta+tantheta|+C#

Rescale #C#:

#I=ln|sqrt61sectheta+sqrt61tantheta|+C#

Reverse the substitution:

#I=ln|x-8+sqrt(x^2-16x+3)|+C#

maganbhai P.
Mar 14, 2018

#I=log|(x-8)+sqrt(x^2-16x+3)|+C#, where,
#C=-logsqrt(61)+c#.
NOTE : -It is better to use #int1/sqrt(X^2+k)dX=ln|x+sqrt(x^2+k)|+c#, from (A)

Explanation:

#I=int1/sqrt(x^2-16x+3)dx=int1/sqrt(x^2-16x+64-61)dx =int1/sqrt((x-8)^2-(sqrt61)^2)dx,.(A)#
We take,#x-8=sqrt(61)sectheta=>x=8+sqrt(61)sectheta=>dx=sqrt(61)secthetatantheta*d(theta)#
#I=int((sqrt(61)secthetatantheta))/sqrt(61sec^2theta-61)d(theta)#
#I=int((sqrt(61)secthetatantheta))/(sqrt(61)sqrt(sec^2theta-1))d(theta)##=int(secthetatantheta)/(tantheta)d(theta)=intsecthetad(theta)##=log|sectheta+tantheta|+c#
#I=log|sectheta+sqrt(sec^2theta-1)|+c#
#I=log|(x-8)/sqrt(61)+sqrt(((x-8)/sqrt61)^2-1)|+c#
#I=log|(x-8)/sqrt(61)+sqrt((x-8)^2-61)/sqrt(61)|+c#
#I=log|(x-8)+sqrt(x^2-16x+3)|-log|sqrt(61)|+c#

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