# How do you integrate int 1/sqrt(x^2-25) by trigonometric substitution?

##### 1 Answer
Dec 19, 2016

${\cosh}^{-} 1 \left(\frac{x}{5}\right) + C$

#### Explanation:

The integrand is defined only for $x \ge 5$.
Substitute $x = 5 \cosh u$, $\mathrm{dx} = 5 \sinh u \mathrm{du}$.
The integral becomes $\int \frac{1}{\sqrt{25 {\cosh}^{2} u - 25}} 5 \sinh u \mathrm{du}$.
Since ${\cosh}^{2} u - {\sinh}^{2} = 1$, this becomes:
$= \int 1 \mathrm{du}$
$= u + C$
$= {\cosh}^{-} 1 \left(\frac{x}{5}\right) + C$
which has an alternative representation in the form of a logarithm.
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