# Integral: sqrt(4-x^2)*sign(x-1) dx

• Alexx1
In summary, the integral given is of the form sqrt(4-x^2)*sign(x-1) dx and can be solved by splitting it into two separate integrals based on the sign function. By using the substitution x=2sint, dx=2costdt and simplifying, the integral can be reduced to 4\int cos^2tdt. Using a double-angle equality for trigonometric functions, the integral can be further simplified to 2t+sin(2t). By substituting t=arcsin(x/2) and simplifying, the solution for the second part can be obtained as (x/2)*sqrt(4-x^2).
Alexx1
Integral: sqrt(4-x^2)*sign(x-1) dx

Can someone help me with this integral?
I've never worked with a sign funtion before, so I have absolutely no idea how to solve this integral

I would be inclined to do this as two separate integrals.

If $x\ge 1$, $x-1\ge 0$ so sng(x-1)= 1 and your integral is $\int \sqrt{4-x^2}dx$.

If x< 1, x-1< 0 so sgn(x-1)= -1 and your integral is $-\int\sqrt{4- x^2}dx$.

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HallsofIvy said:
I would be inclined to do this as two separte integrals.

If $x\ge 1$, $x-1\ge 0$ so sng(x-1)= 1 and your integral is $\int \sqrt{4-x^2}dx$.

If x< 1, x-1< 0 so sgn(x-1)= -1 and your integral is $-\int\sqrt{4- x^2}dx$.

Thanks, but how do you solve $\int \sqrt{4-x^2}dx$
I've tried with: x= 2sin(t) ==> dx= 2cos(t)dt and sin(t)= x/2 ==> t= arcsin(x/2)

But then I get: 2 arcsin(x/2) + sin(2(arcsin(x/2)))
And that's wrong: the first part is correct, but the second part isn't.
What have I done wrong?

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

$x=2sint$
$dx=2costdt$

$$\int \sqrt{4-(2sint)^2}.(2costdt)$$

Simplifying this would give you:

$$4\int cos^2tdt$$

Did you get this? Now, you'll need to use a double-angle equality for the trigo to take the integral of this.

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

$x=2sint$
$dx=2costdt$

$$\int \sqrt{4-(2sint)^2}.(2costdt)$$

Simplifying this would give you:

$$4\int cos^2tdt$$

Did you get this? Now, you'll need to use a double-angle equality for the trigo to take the integral of this.

I solved it till there..
but what I did then was wrong

This is what I did:

$$4\int cos^2tdt$$

= 4((t/2)+(sin(2t)/4))
= 2t + sin (2t)

Because t= arcsin(x/2) you get:

2(arcsin(x/2)) + sin(2arcsin(x/2))

The first part is correct, the second isn't..
What have I done wrong?

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

$x=2sint$
$dx=2costdt$

$$\int \sqrt{4-(2sint)^2}.(2costdt)$$

Simplifying this would give you:

$$4\int cos^2tdt$$

Did you get this? Now, you'll need to use a double-angle equality for the trigo to take the integral of this.

I solved it till there..
but what I did then was wrong

This is what I did:

$$4\int cos^2tdt$$

= 4((t/2)+(sin(2t)/4))
= 2t + sin (2t)

Because t= arcsin(x/2) you get:

2(arcsin(x/2)) + sin(2arcsin(x/2))

The first part is correct, the second isn't..
What have I done wrong?

Alexx1 said:
Because t= arcsin(x/2) you get:

2(arcsin(x/2)) + sin(2arcsin(x/2))

The first part is correct, the second isn't..
What have I done wrong?

Not wrong, just not fully simplified. Often when you have a trig of inverse-trig you can simplify it to an algebraic expression. In this case start out be using :

$$\sin(2 \theta) = 2 \sin(\theta) \, \cos(\theta)$$

and see if you can simplify it from there.

BTW. Since you know it's "wrong" then I assume that you have the correct answer at hand and you know that this second part should simplify to :

$$\frac{x}{2} \, \sqrt{4-x^2}$$

It does. :)

uart said:
Not wrong, just not fully simplified. Often when you have a trig of inverse-trig you can simplify it to an algebraic expression. In this case start out be using :

$$\sin(2 \theta) = 2 \sin(\theta) \, \cos(\theta)$$

and see if you can simplify it from there.

BTW. Since you know it's "wrong" then I assume that you have the correct answer at hand and you know that this second part should simplify to :

$$\frac{x}{2} \, \sqrt{4-x^2}$$

It does. :)

Thank you very much!
But.. when I do this (t= arcsin(x/2)) I get:

sin(2t)
= 2 sin(t) cos (t)
= 2 sin(arcsin(x/2)) cos(arcsin(x/2)
= 2 (x/2) (sqrt(1-(x/2)))

Do I have to bring the 2 into the sqrt to get sqrt(4-x^2)?

Alexx1 said:
Thank you very much!
But.. when I do this (t= arcsin(x/2)) I get:

sin(2t)
= 2 sin(t) cos (t)
= 2 sin(arcsin(x/2)) cos(arcsin(x/2)
= 2 (x/2) (sqrt(1-(x/2)))

Do I have to bring the 2 into the sqrt to get sqrt(4-x^2)?
You are missing the square in $2(x/2)\sqrt{1- (x/2)^2}$.
You don't have to but putting that 2 inside the square root will give you the form of the answer above:
$$(x/2)\sqrt{4(1- x^2/4)}= (x/2)\sqrt{4- x^2}$$

HallsofIvy said:
You are missing the square in $2(x/2)\sqrt{1- (x/2)^2}$.
You don't have to but putting that 2 inside the square root will give you the form of the answer above:
$$(x/2)\sqrt{4(1- x^2/4)}= (x/2)\sqrt{4- x^2}$$

Thank you!

## 1. What is the function being integrated?

The function being integrated is f(x) = √(4-x^2) * sign(x-1), where √(4-x^2) represents the square root of 4-x^2 and sign(x-1) represents the sign function which returns -1 if x-1 is negative, 0 if x-1 is zero, and 1 if x-1 is positive.

## 2. What is the domain of the function?

The domain of the function is all real numbers except for x=1 and values that make the square root of 4-x^2 imaginary. This means that the domain includes all values less than or equal to -2 and greater than or equal to 2.

## 3. What is the meaning of the "dx" at the end of the integral?

The "dx" at the end of the integral represents the variable with respect to which the function is being integrated. In this case, the integral is being taken with respect to x.

## 4. Is there a specific method to solve this integral?

Yes, this integral can be solved using the substitution method. Let u = 4-x^2, then du = -2x dx. By substituting these values and simplifying, the integral becomes ∫ √u * sign(2-x) (-x/2) du. This can be further simplified and solved using basic integration techniques.

## 5. What is the significance of the sign function in this integral?

The sign function in this integral helps to determine the appropriate positive or negative value for the integral as it changes from positive to negative when x=1. This helps to correctly evaluate the integral and obtain the correct result.

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