Integrate ##\int\sqrt{4+x^2} dx##

[chwala]

Homework Statement

$$\int \sqrt{4+x^2}
dx$$

Relevant Equations

hyperbolic equations

still typing...checking latex

 

[chwala]

This is a textbook question...the steps to solution are pretty easy...they indicated use of;

$x=2\sinh u$ that will realize

$$\int \sqrt{4+x^2} dx=\int 2\cosh u⋅ 2\cosh u du=4\cosh^2 u du$$

...from here the steps are pretty clear up to the required solution.

Ok to my question now; Could we also use;

$x=2\cosh u$ instead of $x=2\sinh u$?

Cheers

 

[PeroK]

Ok to my question now; Could we also use;

$x=2\cosh u$ instead of $x=2\sinh u$?

Cheers

That doesn't look so promising to me.

 

[chwala]

That doesn't look so promising to me.

I thought it will be other way round...let me try and see what comes out of it...will share later. Cheers mate.

 

[Orodruin]

Ok to my question now; Could we also use;

$x=2\cosh u$ instead of $x=2\sinh u$?

No. The point is to rewrite $\sqrt{4 + x^2} = \sqrt{4(1+\sinh^2(u))} = 2 \cosh(u)$ using the hyperbolic one. If you would have attempted to use $x = 2 \cosh(u)$ instead you would have ended up with $\sqrt{4 + x^2} = 2\sqrt{1+\cosh^2(u)}$ and $1 + \cosh^2(u)$ does not have any particular simplification in terms of the hyperbolic one. If you would attempt to replace $\cosh^2(u)$ by $\sinh^2(u)$ using the hyperbolic one you would instead end up with $2 + \sinh^2(u)$, which doesn't make you any happier.

 

[Mark44]

Another alternative is to use the substitution $\tan(\theta) = \frac x 2$. This is based on drawing a right triangle with an acute angle $\theta$, where the base is 2 and the side opposite is $x$.

With this substitution, the integral $\int \sqrt{x^2 + 4}~dx$ becomes $4\int \sec^3(\theta)~d\theta$, an integral so well-known there's a wikipedia article devoted to it.

 

[chwala]

Another alternative is to use the substitution $\tan(\theta) = \frac x 2$. This is based on drawing a right triangle with an acute angle $\theta$, where the base is 2 and the side opposite is $x$.

With this substitution, the integral $\int \sqrt{x^2 + 4}~dx$ becomes $4\int \sec^3(\theta)~d\theta$, an integral so well-known there's a wikipedia article devoted to it.

Thanks @Mark44 Let me study on this approach...

 

[Mark44]

The picture I was describing looks like this. The hypotenuse is $\sqrt{x^2 + 4}$.

 

[WWGD]

Its a good rule of thumb to consider trigonometric substitutions when seeing the likes of $\sqrt {a^2 \pm x^2}dx$
When it's a +, consider secant, tangent;
when it's a - , consider sin, cos
Subbing $u=ax$ in each case, as
$1 \pm x^2$ is " translatsble" to either of these.
Edit: Of course, consider the issue of limits of integration.

 

[chwala]

No. The point is to rewrite $\sqrt{4 + x^2} = \sqrt{4(1+\sinh^2(u))} = 2 \cosh(u)$ using the hyperbolic one. If you would have attempted to use $x = 2 \cosh(u)$ instead you would have ended up with $\sqrt{4 + x^2} = 2\sqrt{1+\cosh^2(u)}$ and $1 + \cosh^2(u)$ does not have any particular simplification in terms of the hyperbolic one. If you would attempt to replace $\cosh^2(u)$ by $\sinh^2(u)$ using the hyperbolic one you would instead end up with $2 + \sinh^2(u)$, which doesn't make you any happier.

correct! i just checked...i ended up with,

$$\int \sqrt{(8-4\sinh^2 u)}⋅ 2 \sinh u du$$...

not looking good!