Integrate sqrt ( 4 -x^2)dx

1. Apr 1, 2005

Naeem

Hi,

Can anyone help me integrate :

Integral sqrt ( 4 - x^2) dx

Some ideas:

Make a subsitution x2 = cos (theta)

Taking the derivative of both sides:

2x = - sin (theta).d(theta)

or....

Double angle identity:

(Cosx)^2 = 1 + cos2x / 2

Can anybody give some ideas.

2. Apr 1, 2005

whozum

Trigonometrically: x = 2cos(theta), or 2sin(theta), not $$x^2 = 2cos(theta)$$

You cant use the double angle identity without making that substitution, so its not really 'or' is it :)

You might be able to do it by parts with u = sqrt(4-x^2) and dv = dx.

3. Apr 1, 2005

Jameson

By parts will work.

$$\int udv = uv - \int{vdu}$$

Let u = $\sqrt{4-x^2}$
And du = $$\frac{2x}{2\sqrt{4-x^2}}$$

Let dv = dx
And v = x

$$\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} - \int\frac{2x^2}{2\sqrt{4-x^2}}dx$$

Hope that helps.

4. Apr 1, 2005

HallsofIvy

Staff Emeritus
I don't see any reason for a double angle. If you factor out that 4 you get
$$\sqrt{4- x^2}= 2\sqrt{1- \frac{x^2}{4}$$. And anytime you see something like that you should think "$$\sqrt{1- cos^2x}= sin x$$" (or "$$\sqrt{1- sin^2x}= cos x$$").

Let $$cos(\theta)= \frac{x}{2}$$ or [2 cos(\theta)= x[/tex] so that
$$-2sin(\theta)dx= dx$$. Then $$\integral \sqrt{4-x^2}dx= 2\integral \sqrt{1- (\frac{x}{2})^2}dx= -4\integral\sqrt{1- cos^2\theta}sin \theta d\theta= -4\integral sin^2 \theta d\theta$$.

That last integral you can do using $$sin^2 \theta= \frac{1}{2}(1- cos \theta)$$.
Well, by golly, you do need a double angle!

5. Apr 1, 2005

dextercioby

U missed a "-" (=minus) when computing the marked differential.Not to mention the "dx".

Daniel.

6. Apr 1, 2005

Jameson

You're right. The "dx's" aren't as critical as the minus was... here's the corrected integral.
------------------
$$\int udv = uv - \int{vdu}$$

Let u = $\sqrt{4-x^2}$
And du = $$\frac{-2x}{2\sqrt{4-x^2}}dx$$

Let dv = dx
And v = x

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

Last edited: Apr 1, 2005
7. Apr 1, 2005

dextercioby

Apparently u forgot about it,again.It should have been

$$(...)+\int \frac{x^{2}}{\sqrt{4-x^{2}}} \ dx$$

,where the 2-s got simplified through.

Daniel.

8. Apr 1, 2005

Jameson

Oh, alright. Sorry I didn't see it before. I didn't catch what you were referring to. I'll edit my previous post... hope I get it right this time.

9. Apr 1, 2005

dextercioby

And the last integral can be calculated again by parts...

Daniel.

10. Apr 1, 2005

Jameson

Right. I was hoping the original poster would comment again before I did too many steps.

11. Apr 1, 2005

Naeem

I did this problem using Halls of Ivy idea, i.e substitution, and I got

Integral sqrt( 4 -x^2)dx = -2 Integral ( 1-cos ( theta ) dtheta =
= -2[ Integral dtheta - Integral cos(theta)dtheta ]
= -2 ( theta - sin (theta) + C
= -2 ( cos-1 (x/2) - sin ........ what do I put theta as here)....

12. Apr 1, 2005

Moo Of Doom

You know that $$\sin{\theta}=\sqrt{1-\cos^2{\theta}}$$, right? Does that help?

13. Apr 2, 2005

Naeem

then. theta for the second part would be

theta = arcsin sqrt ( 1- x^2/4).....

14. Apr 2, 2005

whozum

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

$$u = \sqrt{4-x^2}, du = \frac{-x}{\sqrt{4-x^2}}dx, v = x, dv = dx$$

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

x = 2sin(t)
dx = 2cos(t) dt

The integral on the RHS becomes:

$$\int \frac{8cos(t)sin^2(t)}{\sqrt{4-4sin^2(t)}}dt = \int \frac{8cos(t)sin^2(t)}{2\sqrt{cos^2(t)}}dt = \int \frac{8cos(t)sin^2(t)}{2cos(t)}}dt = 4\int{sin^2(t)}{dt}$$

From there use the trig identity$$sin^2(t) = \frac{1-2cos(t)}{2}$$

$$2\int{1-cos(2t)}{dt} = 2t - 2\int{cos2t}{dt} = 2t - sin(2t)$$

From x = 2sin(t), t = arcsin(x/2).

$$2(arcsin(x/2)-sin(2))$$

Since sin(2) is constant it can be omitted and

$$\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + 2arcsin(x/2)$$

15. Apr 2, 2005

Naeem

Is Halls of Ivy method wrong then

16. Apr 2, 2005

dextercioby

Of course not.

Daniel.

17. Apr 3, 2005

whozum

$$\int \sqrt{4-x^2}dx = x\sqrt{4-x^2} + 2arcsin(x/2)$$

When I evaluate the integral on the LHS on maple it tells me that the first term on the RHS should be divided by 2.

Can anyone figure this out?

18. Apr 4, 2005

dextercioby

Maple is correct (up to a constant he never considers ).

$$\int \sqrt{4-x^{2}} \ dx =2 \arcsin\left(\frac{x}{2}\right)+2\frac{x}{2}\sqrt{1-\left(\frac{x}{2}\right)^{2}} + C =2 \arcsin\left(\frac{x}{2}\right)+\frac{1}{2}x\sqrt{4-x^{2}} + C$$

Daniel.

19. Apr 5, 2005

whozum

So you square the 1/2 and put it inside making it 1-(x/2)^2. But why does it still have 4 - x^2?

You lost me.

20. Apr 5, 2005

dextercioby

Well,when reversing the substitution u end up with

$$2\frac{x}{2}\sqrt{1-\left(\frac{x}{2}\right)^{2}}$$

which used to stand for [itex] \sin 2u =2 \sin u \ \cos u [/tex]

U know algebra,so how about work that square & square roots...?

Daniel.