# Homework Help: Integrating the function √(12 -x^2)

1. Oct 21, 2011

### Chrisistaken

1. The problem statement, all variables and given/known data

Integrate the function,

f(x) = √(12 -x2)

2. Relevant equations

n/a

3. The attempt at a solution

I tried splitting the function up as follows:

f(x) = √(12+x)*√(12-x)

then I tried substituting in,

w=12-x and dw=-dx, to get,

∫-2w3/2(2√12-w)1/2dw

and finally I attempted to use integration by parts on the above integral; however this seems to be a dead end. I know the answer is:

1/2*√(-x^2 + 12)*x + 6*arcsin(1/6*sqrt(3)*x)

I just can't seem to work it out for myself.

Any help would be greatly appreciated.

Regards,

Chris

2. Oct 21, 2011

### SammyS

Staff Emeritus
For $\displaystyle \int\sqrt{12-x^2}\,dx$ use the substitution x = (√12)sin(u), then dx = (√12)cos(u) du .

Then $\displaystyle \sqrt{12-x^2}=\sqrt{12-12\sin^2(u)}\,,$ etc.

3. Oct 21, 2011

### Staff: Mentor

As SammyS says, a trig substitution is the way to go here. Since I don't try to remember which trig substitution should be applied, I just sketch a right triangle, labeling the sides and hypotenuse according to the expression inside the radical.

The expression 12 - x2 suggests that the hypotenuse is of length √12, and I can label the altitude as x, and the angle opposite the altitude as θ.

From this I get sin(θ) = x/√12, or x = √12*sin(θ), which is essentially the same as what SammyS is recommending.

4. Oct 22, 2011

### Chrisistaken

Thanks Sammy and Mark,

Will give it a go. Also, I think Mark's the first person I've encountered who's actually given a methodology for determining an appropriate substitution. Thanks :)

Regards,

Chris

5. Oct 22, 2011

### Chrisistaken

Well...

It would seem I have fried my brain with the late nights spent trying to work this out. So far I've got;

∫cos3u.du

and once again, I know the answer (=1/3*sin3u +sinu), but for the life of me I can't figure out how to get there. Is having the function in the form cos3u the right way to go? I swear I'm overlooking something blindingly obvious here.

Regards,

Chris

6. Oct 22, 2011

### Chrisistaken

OK,

Well now I'm trying to use the trig identity cos3x = 4cos3x -3cosx

Sound like the right way to go?

Regards,

Chris

7. Oct 22, 2011

### SammyS

Staff Emeritus
That should work fine, but aren't you integrating cos2(u).

8. Oct 22, 2011

### Chrisistaken

Well now I'm not to sure. After substituting in,

x = √12*sin(u)

I got,

dx = √12*cos(u)du

Putting both of these back into the integral I then had,

∫√(12 -12*sin2(u))*√12*cos(u)du

=∫√(12(1 -sin2(u))*√12*cos(u)du

using the trig identity, cos2(u) +sin2(u) = 1

I then realised whilst writing this that I didn't take the square root of the first cos term. Need sleep.....

Thanks Sammy

Regards,

Chris