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

Jan 11, 2017

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

#### Explanation:

You do not really need a trigonometric substitution here.

If you pose:

$t = {x}^{2} - 7$
$\mathrm{dt} = 2 x \mathrm{dx}$

the integral becomes:

$\int \frac{x \mathrm{dx}}{\sqrt{{x}^{2} - 7}} = \frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{t}} = \sqrt{t} + C = \sqrt{{x}^{2} - 7} + C$

Jan 11, 2017

Although you do not need trigonometric substitution, it can be used for this integral.

#### Explanation:

Recall that ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$, so that's the basic substitution we'll use. (We could use hyperbolic functions instead, but not everyone is familiar with them when first learning trig sub.)

We need, not just ${\sec}^{2} \theta - 7$, but $7 {\sec}^{2} \theta - 7$ so that we can factor out the $7$

$\int \frac{x}{\sqrt{{x}^{2} - 7}} \mathrm{dx}$

Let $x = \sqrt{7} \sec \theta$, so that $\mathrm{dx} = \sqrt{7} \sec \theta \tan \theta d \theta$

and $\sqrt{{x}^{2} - 7} = \sqrt{7 {\sec}^{2} \theta - 7} = \sqrt{7} \tan \theta$

$\int \frac{x}{\sqrt{{x}^{2} - 7}} \mathrm{dx} = \int \frac{\sqrt{7} \sec \theta}{\sqrt{7} \tan \theta} \sqrt{7} \sec \theta \tan \theta d \theta$

$= \int \sqrt{7} {\sec}^{2} \theta d \theta = \sqrt{7} \tan \theta + C$

With $x = \sqrt{7} \sec \theta$, we get $\tan \theta = \frac{\sqrt{{x}^{2} - 7}}{\sqrt{7}}$

Therefore,

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