How do you integrate int 1/sqrt(x^2-4x+13)dx∫1√x2−4x+13dx 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∫1√x2−4x+13=ln∣∣ ∣∣√1+(x−2)29+x−23∣∣ ∣∣+C

Explanation:

int 1/sqrt(x^2-4x+13)d x=int 1/sqrt(x^2-4x+9+4)d x∫1√x2−4x+13dx=∫1√x2−4x+9+4dx
int 1/(sqrt((x-2)^2+3^2))d x∫1√(x−2)2+32dx
x-2=3tan theta" "d x=3sec^2 theta d thetax−2=3tanθ dx=3sec2θdθ
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∫1√x2−4x+13dx=∫3sec2θdθ√9tan2θ+9=∫3sec2θdθ3√1+tan2θ 1+tan2θ=sec2θ
int 1/sqrt(x^2-4x+13)d x=int(3sec^2 theta d theta)/(3sqrt (sec^2 theta))∫1√x2−4x+13dx=∫3sec2θdθ3√sec2θ
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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