# How do you integrate int 1/sqrt(x^2-9x-7)  using trigonometric substitution?

Feb 16, 2016

For this, you're going to have to complete the square.

${x}^{2} - 9 x - 7$

$7 = {x}^{2} - 9 x$

$\frac{28}{4} + \frac{81}{4} = {x}^{2} - 9 x + \frac{81}{4}$

$\frac{109}{4} = {\left(x - \frac{9}{2}\right)}^{2}$

Now bring it back into the problem.

$\int \frac{1}{\sqrt{{\left(x - \frac{9}{2}\right)}^{2} - \frac{109}{4}}} \mathrm{dx}$

Let:
$u = x - \frac{9}{2}$
$\mathrm{du} = \mathrm{dx}$

$= \int \frac{1}{\sqrt{{u}^{2} - \frac{109}{4}}} \mathrm{du}$

Now it looks more like a $\sqrt{{x}^{2} - {a}^{2}}$ form. If we let $a = \frac{\sqrt{109}}{2}$, then:

$u = \frac{\sqrt{109}}{2} \sec \theta$
$\mathrm{du} = \frac{\sqrt{109}}{2} \sec \theta \tan \theta d \theta$
$\sqrt{{u}^{2} - \frac{109}{4}} = \sqrt{\frac{109}{4} {\sec}^{2} \theta - \frac{109}{4}} = \frac{\sqrt{109}}{2} \tan \theta$

What we now have is:

$= \int \frac{1}{\cancel{\frac{\sqrt{109}}{2}} \cancel{\tan \theta}} \cdot \cancel{\frac{\sqrt{109}}{2}} \sec \theta \cancel{\tan \theta} d \theta$

$= \int \sec \theta d \theta$

Remember the trick to do this? See the following:

$\textcolor{g r e e n}{\int \sec x \mathrm{dx}}$

$= \int \frac{\sec x \left(\sec x + \tan x\right)}{\sec x + \tan x} \mathrm{dx}$

$= \int \frac{{\sec}^{2} x + \sec x \tan x}{\sec x + \tan x} \mathrm{dx}$

Let:
$u = \sec x + \tan x$
$\mathrm{du} = \sec x \tan x + {\sec}^{2} x \mathrm{dx}$

$\implies \int \frac{1}{u} \mathrm{du} = \ln | u | + C = \textcolor{b l u e}{\ln | \sec x + \tan x | + C}$

Therefore, what we have is $\ln | \sec \theta + \tan \theta |$ for the result. Now what we can do is substitute in the proper variables.

$\tan \theta = \frac{2 \sqrt{{u}^{2} - \frac{109}{4}}}{\sqrt{109}} = \frac{2 \sqrt{{x}^{2} - 9 x - 7}}{\sqrt{109}}$
$\sec \theta = \left(x - \frac{9}{2}\right) \cdot \frac{2}{\sqrt{109}} = \frac{2 x - 9}{\sqrt{109}}$

$= \textcolor{g r e e n}{\ln | \frac{2 x - 9}{\sqrt{109}} + \frac{2 \sqrt{{x}^{2} - 9 x - 7}}{\sqrt{109}} | + C}$
or, embedding the $\sqrt{109}$ into the $C$:
$= \ln | 2 x - 9 + 2 \sqrt{{x}^{2} - 9 x - 7} | - \ln \sqrt{109} + C$
$= \textcolor{b l u e}{\ln | 2 x - 9 + 2 \sqrt{{x}^{2} - 9 x - 7} | + C}$