integration using arcsin

robertjford80

I see how they get cos^2 = cos 2a + 1

But what I do not understand is how they get

x = 2sin(a)

Jorriss

If this substitution is causing much confusion you could try to integrate by parts. It may at least clear up why there are two terms.

robertjford80

Bohrok,

Thanks for your detailed help. I have never encountered these substitution techniques? Should I have encountered them by now? I'm up to triple integrals in calc. Is one expected to know these techniques by the time they reach triple integrals?

Bohrok

You should see this type of substitution in calc II, definitely before multiple integrals.

Trig substitutions are really a special u-substitution. This problem √(4 - x²), and other similar ones involving square roots and x², could be done with the "u-substitution" u = sin^-1(x/2). Approaching it that way, however, requires that you know how to differentiate inverse trig function, and it may not as clear as thinking of using a slightly different substitution x = 2sin u (or θ as we usually use with trig substitutions) and using trig identities to make the new integrand more manageable.

SammyS

This may seem like beating a dead horse ...

Since you were "having a tough time understanding this step", it seems to me that there's no problem with taking the book's answer & working backwards.

So I took the derivative of [itex]\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ .[/itex]

Here is what it led me to:

Rewrite your integrand, [itex]\displaystyle \sqrt{4-x^2\ }\,,[/itex] as [itex]\displaystyle \frac{4-x^2}{\sqrt{4-x^2\ }}\ .[/itex]

Now split that up into [itex]\displaystyle \frac{4}{\sqrt{4-x^2\ }}-\frac{x^2}{\sqrt{4-x^2\ }}\ .[/itex]

The integral of the first term is [itex]\displaystyle 4\,\sin^{-1}\left(\frac{x}{2}\right)\ .[/itex]

Use integration by parts on the second term, with [itex]u=x[/itex] and [itex]\displaystyle dv=-\frac{x}{\sqrt{4-x^2\ }}dx\,.[/itex]

After this there is one little trick left to finish the problem.

SammyS

Well, the horse appears to have a small amount of life left in it ...

If [itex]\displaystyle dv=-\frac{x}{\sqrt{4-x^2\ }}dx\,,[/itex] then [itex]\displaystyle v=\sqrt{4-x^2\ }\,.[/itex]

So integration by parts gives: [itex]\displaystyle \int{-\frac{x^2}{\sqrt{4-x^2\ }} }dx=x\sqrt{4-x^2\ }-\int{\sqrt{4-x^2\ }}dx\ .[/itex]

Putting this all together gives:

[itex]\displaystyle \int{ \sqrt{4-x^2\ }}\,dx=\int{\frac{4}{\sqrt{4-x^2\ }}}\,dx+\int{-\frac{x^2}{\sqrt{4-x^2\ }} }dx[/itex]

[itex]\displaystyle =4\,\sin^{-1}\left(\frac{x}{2}\right)+x\sqrt{4-x^2\ }-\int{\sqrt{4-x^2\ }}dx\ .[/itex]

Now for that "trick".

Add [itex]\displaystyle \int{\sqrt{4-x^2\ }}dx\,,[/itex] to both sides, divide by 2 ... then multiply by 3 .

dimension10

Let ## x=2sin\theta ##. Then just use the trig identity ##\cos^2\theta=\frac{1+\cos 2\theta}{2}##. Then replace all the thetas with arcsin(x/2) and you are done with the indefinite integral. Then just plug in the limits of the integral and you are done.