Integrating the function √(12 -x^2)

[Chrisistaken]

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,

∫-2w^3/2(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

 

[SammyS]

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.

 

[Mark44]

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.

 

[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

 

[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

 

[Chrisistaken]

OK,

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

Sound like the right way to go?

Regards,

Chris

 

[SammyS]

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).

 

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

Thanks Sammy

Regards,

Chris