How do you integrate #int 1/sqrt(x^2-4x+13)dx# using trigonometric substitution?

2 Answers
ali ergin
Mar 10, 2016

#int 1/sqrt(x^2-4x+13)=l n |sqrt(1+(x-2)^2/9)+(x-2)/3|+C#

Explanation:

#int 1/sqrt(x^2-4x+13)d x=int 1/sqrt(x^2-4x+9+4)d x#
#int 1/(sqrt((x-2)^2+3^2))d x#
#x-2=3tan theta" "d x=3sec^2 theta d theta#
#int 1/sqrt(x^2-4x+13)d x=int (3sec^2 theta d theta)/sqrt(9tan^2 theta+9)=int(3sec^2 theta d theta)/(3sqrt(1+tan^2 theta))" "1+tan^2 theta=sec^2 theta#
#int 1/sqrt(x^2-4x+13)d x=int(3sec^2 theta d theta)/(3sqrt (sec^2 theta))#
#int 1/sqrt(x^2-4x+13)d x=int(cancel(3sec^2 theta) d theta)/(cancel(3sec theta))#
#int 1/sqrt(x^2-4x+13)d x=int sec theta d theta#
#int 1/sqrt(x^2-4x+13)d x=l n|sec theta+tan theta|+C#
#tan theta=(x-2)/3" "sec theta=sqrt(1+tan^2 theta)=sqrt(1+(x-2)^2/9)#
#int 1/sqrt(x^2-4x+13)=l n |sqrt(1+(x-2)^2/9)+(x-2)/3|+C#

trosk
Mar 10, 2016

#sinh^-1((x-2)/3) + C#

Explanation:

The hyperbolic version is also possible:

  • #x-2 = 3 sinh u#
  • #dx = 3 cosh u du#

#int 1/sqrt(x^2-4x+13)dx = int 1/sqrt(9sinh^2 u + 9)3cosh u du = int 1/(3cosh u) 3cosh u du = u + C#

Hence:

#int 1/sqrt(x^2-4x+13)dx = sinh^-1((x-2)/3) + C#

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