Integrate sqrt ( 4 -x^2)dx

[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.

 

[whozum]

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

You can't 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.

 

[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.

 

[HallsofIvy]

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]

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.

 

[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

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

 

[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.

 

[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.

 

[dextercioby]

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

Daniel.

 

[Jameson]

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

 

[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)...

 

[Moo Of Doom]

= -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?

 

[Naeem]

then. theta for the second part would be

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

 

[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

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

 

[Naeem]

Is Halls of Ivy method wrong then

 

[dextercioby]

Of course not.

Daniel.

 

[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?

 

[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.

 

[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.

 

[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]<br /> <br /> U know algebra,so how about work that square & square roots...?<br /> <br /> Daniel.[/itex]

 

[HallsofIvy]

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.

 

[Data]

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 $2\cos \theta = x$.

 

[Data]

oops, didn't notice the second page!

 

[dextercioby]

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

Make the obvious substitution

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

and go from there...

Daniel.

 

[pjkily]

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

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

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

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 please help me? THX!