Question

How do you integrate `int 1/sqrt(x^2+2x)` by trigonometric substitution?

Answer 1

`ln|sqrt((x + 1)^2 - 1) + x + 1| + C`

Explanation:

Complete the square in the denominator (within the `sqrt`).

`int 1/sqrt(1(x^2 + 2x + 1 - 1))dx`

`int 1/sqrt(1(x^2 + 2x + 1) - 1)dx`

`int 1/sqrt((x + 1)^2 - 1)dx`

Let `u = x + 1`. Then `du = dx`.

`int 1/sqrt(u^2 - 1)du`

Now use the substitution `u = sectheta`. Then `du = secthetatanthetad theta`.

`int 1/sqrt(sec^2theta - 1) secthetatantheta d theta`

`int 1/sqrt(tan^2theta) secthetatantheta d theta`

`int 1/tantheta secthetatantheta d theta`

`int sectheta d theta`

This is a known integral that can be derived here

`ln|sectheta + tantheta| + C`

Obviously it's not good enough to stay in `theta`; we have to return to `x`. From our initial substitution, `u/1 = sectheta`. This means that the side opposite `theta` measures `sqrt(u^2 - 1)`. This also means that `tantheta = sqrt(u^2 - 1)/1 = sqrt(u^2 - 1)`.

`ln|sqrt(u^2 - 1) + u| + C`

We have one more substitution to reverse.

`ln|sqrt((x + 1)^2 - 1) + x + 1| + C`

Hopefully this helps!