# How do you integrate int 1/sqrt(t^2-6t+13) by trigonometric substitution?

Dec 20, 2016

${\sinh}^{-} 1 \left(\frac{t - 3}{2}\right) + C$, by completing the square and then either using a hyperbolic substitution, or by citing a standard integral.

#### Explanation:

$= \int \frac{1}{\sqrt{{\left(t - 3\right)}^{2} + 4}} \mathrm{dt}$
which upon substituting $t - 3 = x$ is the standard integral $\int \frac{1}{\sqrt{{x}^{2} + {a}^{2}}} \mathrm{dx} = {\sinh}^{-} 1 \left(\frac{x}{a}\right)$ with $a = 2$.

If the standard integral is not accessible, substitute $x = 2 \sinh u$, $\mathrm{dx} = 2 \cosh u \mathrm{du}$ and use ${\cosh}^{2} u - {\sinh}^{2} u = 1$:

$= \int \frac{2 \cosh u}{\sqrt{4 {\sinh}^{2} u + 4}} \mathrm{du}$
$= \int 1 \mathrm{du}$
$= u + C$
$= {\sinh}^{-} 1 \left(\frac{x}{2}\right) + C$
$= {\sinh}^{-} 1 \left(\frac{t - 3}{2}\right) + C$.

If the $+ 4$ had been $- 4$ you would have substituted $x = 2 \cosh u$.

The final answer can also be expressed using logarithms instead of inverse hyperbolic functions, because ${\sinh}^{-} 1 x = \ln \left(x + \sqrt{{x}^{2} + 1}\right)$, not forgetting also that $\ln \left(\frac{x}{2} + \sqrt{{x}^{2} / 4 + 1}\right)$ differs from $\ln \left(x + \sqrt{{x}^{2} + 4}\right)$ by a constant which can be merged with the arbitrary constant of integration.

You use hyperbolic trig if the quadratic under the square root has a positive coefficient of ${x}^{2}$ and circular trig if it is negative.