Integrating the function √(12 -x^2)

In summary: OK,It would seem that the substitution of x = (√12)sin(u), for x = (√12)cos(u) du, in the function,Integrate the function, f(x) = √(12+x)*√(12-x) gives the correct answer. Once again, thanks to SammyS for providing a methodology for determining an appropriate substitution.
  • #1
Chrisistaken
12
0

Homework Statement



Integrate the function,

f(x) = √(12 -x2)


Homework Equations



n/a

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
 
Physics news on Phys.org
  • #2
For [itex]\displaystyle \int\sqrt{12-x^2}\,dx[/itex] use the substitution x = (√12)sin(u), then dx = (√12)cos(u) du .

Then [itex]\displaystyle \sqrt{12-x^2}=\sqrt{12-12\sin^2(u)}\,, [/itex] etc.
 
  • #3
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
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
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
OK,

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

Sound like the right way to go?

Regards,

Chris
 
  • #7
Chrisistaken said:
OK,

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

Sound like the right way to go?

Regards,

Chris
That should work fine, but aren't you integrating cos2(u).
 
  • #8
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 realized whilst writing this that I didn't take the square root of the first cos term. Need sleep...

Thanks Sammy

Regards,

Chris
 

What is the function √(12 -x^2)?

The function √(12 -x^2) is a mathematical expression that represents the square root of the quantity 12 minus x squared. It is a type of integration problem that involves finding the area under a curve.

What does it mean to integrate a function?

Integrating a function is a mathematical process of finding the area under a curve or the accumulation of a quantity over a given interval. It is the reverse process of differentiation and is used in various fields of science, engineering, and economics.

How do you solve for the integral of √(12 -x^2)?

To solve for the integral of √(12 -x^2), you can use the substitution method or the trigonometric substitution method. Both methods involve substituting variables and simplifying the expression to a form that can be easily integrated. You can also use online tools or software to solve the integral.

What are the applications of integrating the function √(12 -x^2)?

Integrating the function √(12 -x^2) has various applications in physics, engineering, and economics. It can be used to find the area under a curve, calculate the work done by a force, determine the volume of a solid, and calculate the average value of a function over a given interval.

What are some common mistakes when integrating the function √(12 -x^2)?

Some common mistakes when integrating the function √(12 -x^2) include forgetting to apply the chain rule, making incorrect substitutions, and forgetting to add the constant of integration. It is important to carefully follow the steps and double-check your work to avoid making these mistakes.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
269
  • Calculus and Beyond Homework Help
Replies
5
Views
222
  • Calculus and Beyond Homework Help
Replies
9
Views
698
  • Calculus and Beyond Homework Help
Replies
3
Views
267
  • Calculus and Beyond Homework Help
Replies
3
Views
878
  • Calculus and Beyond Homework Help
Replies
19
Views
760
  • Calculus and Beyond Homework Help
Replies
2
Views
479
  • Calculus and Beyond Homework Help
Replies
7
Views
641
  • Calculus and Beyond Homework Help
Replies
7
Views
900
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
Back
Top