How do you integrate #int 1/(x^2sqrt(x^2-7))# by trigonometric substitution?

1 Answer
Nov 25, 2016

#sqrt(x^2-7)/(7x)+C#

Explanation:

#I=intdx/(x^2sqrt(x^2-7))#

Apply the substitution #x=sqrt7sectheta,dx=sqrt7secthetatanthetad theta#.

#color(white)I=int(sqrt7secthetatanthetad theta)/(7sec^2thetasqrt(7sec^2theta-7))#

#color(white)I=1/7int(tanthetad theta)/(secthetasqrt(sec^2theta-1))#

#color(white)I=1/7int(d theta)/sectheta#

#color(white)I=1/7intcosthetad theta#

#color(white)I=1/7sintheta#

#color(white)I=1/7tantheta/sectheta#

#color(white)I=1/7sqrt(sec^2theta-1)/sectheta#

Using #x=sqrt7sectheta=>sectheta=x/sqrt7#:

#color(white)I=1/7sqrt(x^2/7-1)/(x/sqrt7)#

#color(white)I=1/7(1/sqrt7sqrt(x^2-7))/(x/sqrt7)#

#color(white)I=sqrt(x^2-7)/(7x)+C#