How do you integrate #int 1/(x^2sqrt(x^2-7))# by trigonometric substitution?
1 Answer
Nov 25, 2016
Explanation:
#I=intdx/(x^2sqrt(x^2-7))#
Apply the substitution
#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
#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#