integration using arcsin

robertjford80

1. The problem statement, all variables and given/known data





3. The attempt at a solution

I'm having a tough time understanding this step. I understand that

the integral of 1/sqrt(a^2-x^2) = arcsin(u/a) + C

But sqrt(4-x^2) does not have a 1 in the numerator. It looks like they're using u substitution, if so, then why is it + 4 arcsin rather than times 4arcsin. I also don't understand where the 4 comes from.

spamiam

I'm guessing it's just a typical trig sub. Set ##x = 2 \sin \theta##, use a trigonometric identity, integrate and then solve for ##\theta## in terms of x.

If you haven't seen this technique before, check out this link.

SammyS

What's the derivative of

[itex]\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ ?[/itex]

robertjford80

If I knew I wouldn't have posted the thread

robertjford80

well, if If the integrand contains x^2 − a^2, let

[tex] x = a \sec \theta\,[/tex]

and use the identity

[tex] \sec^2 \theta-1 = \tan^2 \theta.\,[/tex]

Then why do they use arcsin instead of secant

SammyS

How do you expect to be able to integrate, if you don't know basic differentiation ?

According to your OP, the derivative of [itex]\displaystyle \sin^{-1}\left(\frac{x}{2} \right) [/itex] is [itex]\displaystyle \frac{1}{\sqrt{4-x^2}}\ .[/itex]

The rest is pretty basic.

Bohrok

There's a very important difference in whether you have x² - a² or a² - x² so that you use x = asecθ or x = asinθ, respectively.

robertjford80

well, is a^2-x^2 which results in asin. However, that integral only works if there is a one in the numerator and clearly there is not, so I'm still wondering where 4asin came from.

Sammy, the rest is not basic.

SammyS

How is [itex]\displaystyle \frac{d}{dx}\left(x\sqrt{4-x^2\,}\right)[/itex] not basic ?

robertjford80

It's basic alright, but it doesn't even appear on the left side of the equation. Why is the text using that derivative?

Bohrok

∫√(4 - x²) dx

x = 2sinθ → x² = 2²sin²θ
dx = 2cosθ dθ

[tex]\int\sqrt{4-4\sin^2\theta}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4(1-sin^2\theta)}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4\cos^2\theta}\text{ }2\cos\theta\text{ }d\theta[/tex][tex]= \int 2\cos\theta\text{ }2\cos\theta\text{ }d\theta = 4\int \cos^2\theta\text{ }d\theta = 4\int\frac{1}{2}(\cos2\theta + 1)\text{ }d\theta[/tex]The cos2θ part will turn into x√(4 - x²) and the 1 part will turn into sin^-1(x/2) after substituting back for x and simplifying.

spamiam

robert, the method I outlined in my original response (which I've quoted above) is still correct. You had the right idea, but as Bohrok pointed out, the substitution should be ##x = 2 \sin \theta## (as I said originally). You can then use the identity ##1 - \sin^2 \theta = \cos^2 \theta##. What does the integral become with this substitution?

Edit: Bohrok seems to have done a lot of the work for you.

Note that he or she did leave out the factor of 3, however.

SammyS

The reason I asked this question is that this derivative should be equal to [itex]\displaystyle 3\sqrt{4-x^2}\ .[/itex]

The steps involved in taking this derivative might help you understand how one finds the anti-derivative.

robertjford80

How is this legal?

By what rule? I don't see how you can get cos to turn into x

robertjford80

The book says they're equal so I already know that.

That's what I'm trying to figure out.

spamiam

Legal? It's a substitution. If I said "Let ##\theta = \arcsin\left(x/2\right)##," would you still take issue with it? Certainly there is a lot of work you'd have to do to provide a rigorous proof that it works, but I'm guessing you've done substitutions before without taking a real analysis course. Take a look at the link I posted above.

The last step that Bohrok did uses a double-angle formula. See if you can work it out. Once you've integrated (in terms of ##\theta##) substitute back in for x.

Bohrok

We're driven to use x = 2sinθ since the domain of √(4 - x²) is [-2, 2], and the range of 2sinθ is [-2, 2]. Not tired just yet, so I typed up the steps (ignoring +C)

[tex]x = 2\sin\theta \Longrightarrow \frac{x}{2} = \sin\theta \Longrightarrow\sin^{-1}\left(\frac{x}{2}\right) = \theta[/tex][tex]\int\cos2\theta\text{ }d\theta = \frac{1}{2}\sin2\theta = \frac{1}{2}\cdot2\sin\theta\cos\theta = \sin\theta\cos\theta = \frac{x}{2}\cos\left(\sin^{-1}\frac{x}{2}\right) = \frac{x}{2}\sqrt{1 - \left(\frac{x}{2}\right)^2} = \frac{x}{2}\sqrt{1 - \frac{x^2}{4}}[/tex]
There's a trig identity up there where cos(sin^-1x) = √(1 - x²)