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

[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

 

[HallsofIvy]

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

 

[Alexx1]

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?

 

[Mentallic]

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

Now, using your substitutions...

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

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

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.

 

[Alexx1]

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

Now, using your substitutions...

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

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

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]

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

Now, using your substitutions...

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

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

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?

 

[uart]

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. :)

 

[Alexx1]

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)?

 

[HallsofIvy]

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:
$$(x/2)\sqrt{4(1- x^2/4)}= (x/2)\sqrt{4- x^2}$$

 

[Alexx1]

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!