Sin h to the Negative Integral

[Jason76]

Post Edited With Correct Answer

\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + 1}} = sin h^{-1} u + C\)

Is there anyway to spot a situation when we might use this (or for that matter similar integral formulas)?

Problem:

\(\displaystyle \int \dfrac{dx}{\sqrt{9x^{2} + 4}}\)

The denominator is a square root, 4 in the problem above is easily reduced to one, and the "u" term is a square so we would use (unless we want to use a "tan method"):

\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + 1}} = sin h^{-1} u + C\)

\(\displaystyle \int \dfrac{dx}{\sqrt{9x^{2} + 4}}\)

Multiply by \(\displaystyle \dfrac{1}{4}\) to get it into form.

\(\displaystyle \int \dfrac{dx}{\sqrt{(3x/2)^{2} + 1}}\)

\(\displaystyle u = (\dfrac{3x}{2})^{2}\)

\(\displaystyle du = \dfrac{3}{2} dx\)

\(\displaystyle du \dfrac{2}{3} = dx\) isolating dx

Two constants of integration come up: \(\displaystyle \dfrac{1}{2}\) (from taking the square root of \(\displaystyle \dfrac{1}{4}\)) and \(\displaystyle \dfrac{2}{3}\) (from regular substitution).

\(\displaystyle (\dfrac{1}{2}) (\dfrac{2}{3}) \int \dfrac{dx}{\sqrt{(3x/2)^{2} + 1}}\)

\(\displaystyle \dfrac{1}{3} \int \dfrac{3/2 dx}{\sqrt{(3x/2)^{2} + 1}}\)

\(\displaystyle \dfrac{1}{3} sin h^{-1} \dfrac{3x}{2} + C\) - Final Answer

 

[Deleted member 4993]

\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + 1}}= sin h^{-1} u + C\)

Is there anyway to spot a situation when we might use this (or for that matter similar integral formulas)?

Problem:

\(\displaystyle \dfrac{dx}{\sqrt{9x^{2} + 4}}\)

Substitute:

u = 3/2 * x

 

[Jason76]

Question Answered in Original Edited Post

I think the 1/3 constant of integration comes from the product of two constants of integration \(\displaystyle \dfrac{2}{3}\) (from the denominator) and \(\displaystyle \dfrac{1}{2}\)(from the 1/4 that was used to get the equation in proper form) Can anyone confirm this?

Also, one way to know that using the "sin h" method is the route to go is: You can't factor it, and you can't complete the square on it. If you could then the "cos h" method might work. But for both methods a square root is needed in the denominator.

 

[Deleted member 4993]

\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + 1}} = \sin h^{-1} u + C\)

Is there anyway to spot a situation when we might use this (or for that matter similar integral formulas)?

It is not necessary to use sinh^-1 form

substitute:

u = tan(Θ)

du = sec²(Θ) dΘ

then:

\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + 1}} \)

\(\displaystyle =\ \int \dfrac{sec^2(\theta) \ d\theta}{\sqrt{tan^{2}(\theta) + 1}} \)

\(\displaystyle =\ \int \sec(\theta) \ d\theta \)

= ln|sec(Θ) + tan(Θ)| + C

Now substitute back.......

 

[Jason76]

It is not necessary to use sinh^-1 form

substitute:

u = tan(Θ)

du = sec²(Θ) dΘ

then:

\(\displaystyle \int \dfrac{du}{\sqrt{u^{2} + 1}} \)

\(\displaystyle =\ \int \dfrac{sec^2(\theta) \ d\theta}{\sqrt{tan^{2}(\theta) + 1}} \)

\(\displaystyle =\ \int \sec(\theta) \ d\theta \)

= ln|sec(Θ) + tan(Θ)| + C

Now substitute back.......

Either method is equally easy.

Tracking back on what I said, you can simply look at the appearance of the integral formula for deciding which one to use. In this problem's case, we see it's easily reduced to 1, so the "tan method" or "sin h" methods are applicable.