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# How do you integrate int 1/sqrt(x^2-16x-7)  using trigonometric substitution?

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Jun 18, 2018

#### Answer:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \ln \left\mid \left(x - 8\right) + \sqrt{{x}^{2} - 16 x - 7} \right\mid + C$

#### Explanation:

Complete the square under the root:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \int \frac{\mathrm{dx}}{\sqrt{{\left(x - 8\right)}^{2} - 73}}$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \frac{1}{\sqrt{73}} \int \frac{\mathrm{dx}}{\sqrt{{\left(\frac{x - 8}{\sqrt{73}}\right)}^{2} - 1}}$

Substitute:

$u = \frac{x - 8}{\sqrt{73}}$

$\mathrm{du} = \frac{\mathrm{dx}}{\sqrt{73}}$

so:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \int \frac{\mathrm{du}}{\sqrt{{u}^{2} - 1}}$

Restrict now to the interval $u \in \left(1 , + \infty\right)$ and substitute:

$u = \cosh t$

$\mathrm{du} = \sinh t \mathrm{dt}$

with $t \in \left(0 , + \infty\right)$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \int \frac{\sinh t \mathrm{dt}}{\sqrt{{\cosh}^{2} t - 1}}$

As ${\cosh}^{2} t - {\sinh}^{2} t = 1$ and $\sinh t > 0$ for $t > 0$:

$\sqrt{{\cosh}^{2} t - 1} = \sinh t$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \int \frac{\sinh t \mathrm{dt}}{\sinh} t = \int \mathrm{dt} = t + C$

undoing the substitution:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \text{arccosh} \left(\frac{x - 8}{\sqrt{73}}\right) + C$

and using the logarithmic form of the inverse hyperbolic cosine:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \ln \left\mid \frac{x - 8}{\sqrt{73}} + \sqrt{{\left(\frac{x - 8}{\sqrt{73}}\right)}^{2} - 1} \right\mid + C$

Multiplying the argument of the logarithm by $\sqrt{73}$ only changes the constant $C$, so:

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \ln \left\mid \left(x - 8\right) + \sqrt{{\left(x - 8\right)}^{2} - 73} \right\mid + C$

$\int \frac{\mathrm{dx}}{\sqrt{{x}^{2} - 16 x - 7}} = \ln \left\mid \left(x - 8\right) + \sqrt{{x}^{2} - 16 x - 7} \right\mid + C$

and by differentiating we can verify that the solution is valid also for $x - 8 < - \sqrt{73}$

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