- #1

Orion1

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How is this problem integrated?

[tex]\int \sqrt{ \sin x} \; dx[/tex]

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- Thread starter Orion1
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- #1

Orion1

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How is this problem integrated?

[tex]\int \sqrt{ \sin x} \; dx[/tex]

- #2

Zurtex

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The solution can not be given in terms of elementary functions.

- #3

lurflurf

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No elementary functions have [tex]\sqrt{\sin(x)}[/tex] as their derivative. It looks like it an antiderivative could be expressed using elliptic integral of the second kind.

http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html

http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html

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- #4

Zurtex

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Indeed, according to mathematica where:lurflurf said:No elementary functions have [tex]\sqrt{\sin(x)}[/tex] as their derivative. It looks like it an antiderivative could be expressed using elliptic integral of the second kind.

http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html

[tex]\text{EllipticE} (z,m) = \int_0^z \sqrt{1 - m \sin^2 t} \, dt[/tex]

Then:

[tex]\int \sqrt{ \sin x } \, dx = -2 \text{EllipticE} \left( \frac{1}{4} ( \pi - 2x ), 2 \right)[/tex]

- #5

saltydog

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Zurtex said:Indeed, according to mathematica where:

[tex]\text{EllipticE} (z,m) = \int_0^z \sqrt{1 - m \sin^2 t} \, dt[/tex]

Then:

[tex]\int \sqrt{ \sin x } \, dx = -2 \text{EllipticE} \left( \frac{1}{4} ( \pi - 2x ), 2 \right)[/tex]

Alright, I had problems with it:

Show:

[tex]\int \sqrt{ \sin x } \, dx = -2 \text{EllipticE} \left( \frac{1}{4} ( \pi - 2x ), 2 \right)[/tex]

So differentiating the RHS:

[tex]\frac{d}{dx}\left(-2\int_0^{\frac{\pi-2x}{4}} \sqrt{1-2Sin^2(t)}dt\right)=\sqrt{1-2Sin^2(\frac{\pi}{4}-\frac{x}{2})}[/tex]

Well anyway, I'll save the rest for others to go through if they need practice like me to show that this is equal to:

[tex]\sqrt{\sin x}[/tex]

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- #6

Zurtex

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I had the same the same odd feeling about it when I looked at it and enjoyed doing my old A-Level work on proving trigonometric identities.saltydog said:Well anyway, I'll save the rest for others to go through if they need practice like me to show that this is equal to:

[tex]\sqrt{\sin x}[/tex]

- #7

Orion1

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Integrating with EllipticE formula:

[tex]\int \sqrt{\cos x} \; dx = 2 \int_0^{\frac{x}{2}} \sqrt{1 - 2 \sin^2 (t)} \; dt[/tex]

RHS Differentiation:

[tex]\frac{d}{dx} \left( 2 \int_0^{\frac{x}{2}} \sqrt{1 - 2 \sin^2 (t)} \; dt \right) = \sqrt{1 - 2 \sin^2 \left( \frac{x}{2} \right)}[/tex]

Trigonometric Identity:

[tex]1 - 2 \sin^2 \left( \frac{x}{2} \right) = \cos x[/tex]

Any Calculus I students interested in integrating this formula?

[tex]\int \sqrt{ \tan x} \; dx[/tex]

- #8

Zurtex

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Orion1 said:

Any Calculus I students interested in integrating this formula?

[tex]\int \sqrt{ \tan x} \; dx[/tex]

Haha, that's really nasty, if anyone wants a go at this then I'll just help out by saying that you don't need to put it of the form of an EllipticE formula

- #9

lurflurf

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It is not that bad. It should probably be on a list of good calculus final questions along with.Zurtex said:Haha, that's really nasty, if anyone wants a go at this then I'll just help out by saying that you don't need to put it of the form of an EllipticE formula

[tex]\frac{d}{dx}x^x[/tex]

and

[tex]\int x^3e^{-2x}\sin(x)dx[/tex]

- #10

Zurtex

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I got asked that question on my university interview, one of the few people who did it without any helplurflurf said:It is not that bad. It should probably be on a list of good calculus final questions along with.

[tex]\frac{d}{dx}x^x[/tex]

- #11

TD

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The following substitutionOrion1 said:Any Calculus I students interested in integrating this formula?

[tex]\int \sqrt{ \tan x} \; dx[/tex]

[tex]\tan x = y^2 \Leftrightarrow x = \arctan \left( {y^2 } \right) \Leftrightarrow dx = \frac{{2y}}{{y^4 + 1}}dy[/tex]

gives: [tex]\int {\frac{{2y^2 }}{{y^4 + 1}}dy}[/tex]

That should be doable

- #12

lurflurf

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The only potential problems is that if one "refuses" to use complex numbers, effecting the integration requires some rather unmotivated ad hoc manipulations. Probably some rather clever trig identitiy manipulation would get the job done as well.TD said:The following substitution

[tex]\tan x = y^2 \Leftrightarrow x = \arctan \left( {y^2 } \right) \Leftrightarrow dx = \frac{{2y}}{{y^4 + 1}}dy[/tex]

gives: [tex]\int {\frac{{2y^2 }}{{y^4 + 1}}dy}[/tex]

That should be doable

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- #13

Orion1

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Have any Calculus I students attempted to compute this formula on a >TI-89 calculator?

[tex]\int \sqrt{ \tan x} \; dx[/tex]

How long did the computation take? stopwatch?

Now compute this formula:

[tex]\int {\frac{{2y^2 }}{{y^4 + 1}}dy}[/tex]

How long did the computation take?

Have any Calculus I students compared the tangent formula solution generated from a >TI-89 as compared to the Mathematica solution?

Reference:

http://integrals.wolfram.com/

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