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

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mason m Share
Sep 12, 2016

$\ln \left\mid x + \sqrt{{x}^{2} - {a}^{2}} \right\mid + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - {a}^{2}}}$

We will use the substitution $x = a \sec \theta$. Thus $\mathrm{dx} = a \sec \theta \tan \theta d \theta$. Substituting:

$= \int \frac{a \sec \theta \tan \theta d \theta}{\sqrt{{a}^{2} {\sec}^{2} \theta - {a}^{2}}} = \int \frac{a \sec \theta \tan \theta d \theta}{a \sqrt{{\sec}^{2} \theta - 1}}$

Note that ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$:

$= \int \frac{\sec \theta \tan \theta d \theta}{\tan} \theta = \int \sec \theta d \theta = \ln \left\mid \sec \theta + \tan \theta \right\mid + C$

From $x = a \sec \theta$ we see that $\sec \theta = \frac{x}{a}$. Thus we have a right triangle where $x$ is the hypotenuse and $a$ is the adjacent side. Through the Pythagorean theorem we see that the opposite side is $\sqrt{{x}^{2} - {a}^{2}}$.

So, $\tan \theta$ would be opposite over adjacent, or $\frac{\sqrt{{x}^{2} - {a}^{2}}}{a}$.

$= \ln \left\mid \frac{x}{a} + \frac{\sqrt{{x}^{2} - {a}^{2}}}{a} \right\mid + C$

Note that a $\frac{1}{a}$ term can be factored from both of these, which can then be removed from the logarithm as a constant: $\log \left(A B\right) = \log \left(A\right) + \log \left(B\right)$. The $\ln \left(\frac{1}{a}\right)$ constant will be absorbed into $C$.

$= \ln \left\mid x + \sqrt{{x}^{2} - {a}^{2}} \right\mid + C$

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