Integration with variable substitution

[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.

 

[PeroK]

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

 

[Rectifier]

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

 

[PeroK]

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

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

 

[Rectifier]

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

 

[PeroK]

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

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?

 

[ehild]

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 $t= \sqrt{a+x^2}$

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

Spoiler

Use the substitution √a sinh(u)=x

 

[Ray Vickson]

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.

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.