# Homework Help: Integration using arcsin

1. May 27, 2012

### 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.

2. May 27, 2012

### 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.

3. May 27, 2012

### SammyS

Staff Emeritus
What's the derivative of
$\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ ?$​

Last edited: May 28, 2012
4. May 27, 2012

### robertjford80

If I knew I wouldn't have posted the thread

5. May 27, 2012

### robertjford80

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

$$x = a \sec \theta\,$$

and use the identity

$$\sec^2 \theta-1 = \tan^2 \theta.\,$$

Then why do they use arcsin instead of secant

6. May 27, 2012

### SammyS

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

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

The rest is pretty basic.

7. May 27, 2012

### 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.

8. May 28, 2012

### 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.

9. May 28, 2012

### SammyS

Staff Emeritus
How is $\displaystyle \frac{d}{dx}\left(x\sqrt{4-x^2\,}\right)$ not basic ?

10. May 28, 2012

### 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?

11. May 28, 2012

### Bohrok

∫√(4 - x2) dx

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

$$\int\sqrt{4-4\sin^2\theta}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4(1-sin^2\theta)}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4\cos^2\theta}\text{ }2\cos\theta\text{ }d\theta$$$$= \int 2\cos\theta\text{ }2\cos\theta\text{ }d\theta = 4\int \cos^2\theta\text{ }d\theta = 4\int\frac{1}{2}(\cos2\theta + 1)\text{ }d\theta$$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.

12. May 28, 2012

### 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. :grumpy:

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

13. May 28, 2012

### SammyS

Staff Emeritus
The reason I asked this question is that this derivative should be equal to $\displaystyle 3\sqrt{4-x^2}\ .$

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

14. May 28, 2012

### robertjford80

How is this legal?

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

15. May 28, 2012

### robertjford80

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

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

16. May 28, 2012

### 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.

Last edited: May 28, 2012
17. May 28, 2012

### 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)

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

18. May 28, 2012

### robertjford80

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

But what I do not understand is how they get

x = 2sin(a)

19. May 28, 2012

### 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.

20. May 28, 2012

### 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?

21. May 28, 2012

### 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.

22. May 28, 2012

### SammyS

Staff Emeritus
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 $\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ .$

Here is what it led me to:
Rewrite your integrand, $\displaystyle \sqrt{4-x^2\ }\,,$ as $\displaystyle \frac{4-x^2}{\sqrt{4-x^2\ }}\ .$

Now split that up into $\displaystyle \frac{4}{\sqrt{4-x^2\ }}-\frac{x^2}{\sqrt{4-x^2\ }}\ .$

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

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

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

23. May 29, 2012

### SammyS

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

If $\displaystyle dv=-\frac{x}{\sqrt{4-x^2\ }}dx\,,$ then $\displaystyle v=\sqrt{4-x^2\ }\,.$

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

Putting this all together gives:
$\displaystyle \int{ \sqrt{4-x^2\ }}\,dx=\int{\frac{4}{\sqrt{4-x^2\ }}}\,dx+\int{-\frac{x^2}{\sqrt{4-x^2\ }} }dx$
$\displaystyle =4\,\sin^{-1}\left(\frac{x}{2}\right)+x\sqrt{4-x^2\ }-\int{\sqrt{4-x^2\ }}dx\ .$​

Now for that "trick".

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

24. May 29, 2012

### 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.