# Integration with variable substitution

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1. Dec 21, 2016

### Rectifier

Hello, I am having trouble with solving the problem below

The problem

Find all primitive functions to $f(x) = \frac{1}{\sqrt{a+x^2}}$.
(Translated to English)

The attempt
I am starting with substituting $t= \sqrt{a+x^2} \Rightarrow x = \sqrt{t^2 - a}$ in $$\int \frac{1}{\sqrt{a+x^2}} \ dx \rightarrow \int \frac{1}{t} \ dx$$
$\frac{dx}{dt} = D \left( \sqrt{t^2 - a} \right) = \frac{1}{2 \sqrt{t^2 - a}} \cdot 2t = \frac{t}{\sqrt{t^2 - a}}$
.
$dx = \frac{t}{\sqrt{t^2 - a}} \cdot dt$

Together we get:
$$\int \frac{1}{t} \frac{t}{\sqrt{t^2 - a}} \cdot dt = \int \frac{1}{\sqrt{t^2 - a}} \cdot dt$$

This is where I get stuck since this integral is not much easier to solve that the original.

2. Dec 21, 2016

### PeroK

With these sorts of integrals, you should be thinking about a trig substitution.

3. Dec 21, 2016

### Rectifier

But the answer to this problem is $\ln|x + \sqrt{x^2+a}| + C$ and not a trig function.

4. Dec 21, 2016

### PeroK

Ha! You might be better off in this case not knowing the answer. That is actually a trig function in disguise!

5. Dec 21, 2016

### Rectifier

Unfortunately, that does not make my life easier :,(

6. Dec 21, 2016

### PeroK

Let's assume $a$ is positive. And, let $x = \sqrt{a} f(u)$, where $f$ is some unknown trig function, then we need to look at:

$x^2 + a = a(f^2(u) + 1)$ and $dx = \sqrt{a} f'(u)du$

You're looking for a trig function where $f^2(u) + 1$ is an identity. The obvious one is $f(u) = \tan(u)$. You can try that and you should (eventually) get the answer. But, perhaps there's a better one to try?

7. Dec 21, 2016

### ehild

Think of the hyperbolic functions, and the most basic relationship between cosh and sinh.

Use the substitution √a sinh(u)=x

Last edited: Dec 21, 2016
8. Dec 21, 2016

### Ray Vickson

If $a > 0$ you can write the integrand as $1/\sqrt{x^2+b^2}$. Now the identity $b^2 \cosh^2(t) = b^2 \sinh^2(t) + b^2$ comes to mind. If $a < 0$ you can write the integrand as $1/\sqrt{x^2 - b^2}$, and the identity $b^2 \sinh^2(t) = b^2 \cosh^2(t) - b^2$ cones to mind.