Question

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

Answer 1

`int (xdx)/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`
`dt = 2xdx`

the integral becomes:

`int (xdx)/sqrt(x^2-7) = 1/2 int (dt)/sqrt(t) = sqrt(t) + C = sqrt(x^2-7)+C`

Answer 2

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

Explanation:

Recall that `sec^2theta-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 `7sec^2 theta - 7` so that we can factor out the `7`

`int x/sqrt(x^2-7) dx`

Let `x = sqrt7 sec theta`, so that `dx = sqrt7 sec theta tan theta d theta`

and `sqrt(x^2-7) = sqrt (7sec^2 theta - 7) = sqrt7 tan theta`

`int x/sqrt(x^2-7) dx = int (sqrt7sec theta)/(sqrt7 tan theta) sqrt7 sec theta tan theta d theta`

`= int sqrt7 sec^2 theta d theta = sqrt7 tan theta +C`

With `x = sqrt7 sec theta`, we get `tan theta = sqrt(x^2-7)/sqrt7`

Therefore,

`int x/sqrt(x^2-7) dx = sqrt(x^2-7) +C`