- #1

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__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

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- #1

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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

- #2

HallsofIvy

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

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

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

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

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

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- #3

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Thanks, but how do you solve [itex]\int \sqrt{4-x^2}dx[/itex]I would be inclined to do this as two separte integrals.

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

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

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?

- #4

Mentallic

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Now, using your substitutions...

[itex]x=2sint[/itex]

[itex]dx=2costdt[/itex]

[tex]\int \sqrt{4-(2sint)^2}.(2costdt)[/tex]

Simplifying this would give you:

[tex]4\int cos^2tdt[/tex]

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

- #5

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I solved it till there..

Now, using your substitutions...

[itex]x=2sint[/itex]

[itex]dx=2costdt[/itex]

[tex]\int \sqrt{4-(2sint)^2}.(2costdt)[/tex]

Simplifying this would give you:

[tex]4\int cos^2tdt[/tex]

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

but what I did then was wrong

This is what I did:

[tex]4\int cos^2tdt[/tex]

= 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?

- #6

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I solved it till there..

Now, using your substitutions...

[itex]x=2sint[/itex]

[itex]dx=2costdt[/itex]

[tex]\int \sqrt{4-(2sint)^2}.(2costdt)[/tex]

Simplifying this would give you:

[tex]4\int cos^2tdt[/tex]

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

but what I did then was wrong

This is what I did:

[tex]4\int cos^2tdt[/tex]

= 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?

- #7

uart

Science Advisor

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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 :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?

[tex] \sin(2 \theta) = 2 \sin(\theta) \, \cos(\theta)[/tex]

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 :

[tex] \frac{x}{2} \, \sqrt{4-x^2} [/tex]

It does. :)

- #8

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Thank you very much!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 :

[tex] \sin(2 \theta) = 2 \sin(\theta) \, \cos(\theta)[/tex]

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 :

[tex] \frac{x}{2} \, \sqrt{4-x^2} [/tex]

It does. :)

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)?

- #9

HallsofIvy

Science Advisor

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You are missing the square in [itex]2(x/2)\sqrt{1- (x/2)^2}[/itex].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 don't

[tex](x/2)\sqrt{4(1- x^2/4)}= (x/2)\sqrt{4- x^2}[/tex]

- #10

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Thank you!You are missing the square in [itex]2(x/2)\sqrt{1- (x/2)^2}[/itex].

You don'thaveto but putting that 2 inside the square root will give you the form of the answer above:

[tex](x/2)\sqrt{4(1- x^2/4)}= (x/2)\sqrt{4- x^2}[/tex]

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