# Integrate sqrt ( 4 -x^2)dx

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.

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.

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.

HallsofIvy
Homework Helper
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!

dextercioby
Homework Helper
Jameson said:
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.
U missed a "-" (=minus) when computing the marked differential.Not to mention the "dx".

Daniel.

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

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

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dextercioby
Homework Helper
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.

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.

dextercioby
Homework Helper
And the last integral can be calculated again by parts...

Daniel.

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

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)....

Naeem said:
= -2 ( cos-1 (x/2) - sin ........ what do I put theta as here)....
You know that $$\sin{\theta}=\sqrt{1-\cos^2{\theta}}$$, right? Does that help?

then. theta for the second part would be

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

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

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

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

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

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

Is Halls of Ivy method wrong then

dextercioby
Homework Helper
Of course not.

Daniel.

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

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?

dextercioby
Homework Helper
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.

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.

dextercioby
Homework Helper
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 $\sin 2u =2 \sin u \ \cos u [/tex] U know algebra,so how about work that square & square roots...? Daniel. HallsofIvy Science Advisor Homework Helper The fact that there are different ways of doing a problem does not mean that any of them is wrong! The original problem was Integral sqrt ( 4 - x^2) dx $$\integral \sqrt{4-x^2}dx$$ Let $$2 sin(\theta)= x$$ so that 2 cos(\theta)= dx. $$\sqrt{4- x^2}= \sqrt{4- 4sin^2(\theta)}= 2 cos(\theta)$$. The integral becomes $$4 \int cos^2(\theta)d\theta= 2 \int (1+ cos(2\theta))d\theta = 2\theta+ sin(2\theta)$$. Now, go back to x. Last edited by a moderator: No, it's fine. Here: $$\int \sqrt{4-x^2} \ dx = \int -2\sin \theta \sqrt{4 - 4\cos^2 \theta} \ dx = -4 \int \sin^2 \theta \ dx = -2\int 1 - \cos 2\theta \ d\theta = -2\theta + \sin 2\theta + C = -2\arccos \left( \frac{x}{2} \right) + 2\sin \theta \cos \theta+C$$ $$= \frac{x}{2}\sqrt{4-x^2} - 2\arccos \left( \frac{x}{2} \right)+C$$ this is an equivalent answer (it differs only from those above by a constant). Note that here [itex]2\cos \theta = x$.

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oops, didn't notice the second page!

dextercioby
Homework Helper
Here's another way to do it,if u have the patience (:tongue2:).A mathematician should.

Make the obvious substitution

$$\frac{x}{2}=\tanh u$$

and go from there...

Daniel.

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

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

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

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

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

at the very end, how come sin2t becomes a sin2?
If you substitute t=arcsin(x/2) into sin2t, don't we get sin(2arcsin(x/2))?
how could that come out to be sin2?

Can someone plz help me? THX!!!