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Orion1
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How is this problem integrated?
[tex]\int \sqrt{ \sin x} \; dx[/tex]
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
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]
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]
Orion1 said:
Any Calculus I students interested in integrating this formula?
[tex]\int \sqrt{ \tan x} \; dx[/tex]
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
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]
The following substitutionOrion1 said:Any Calculus I students interested in integrating this formula?
[tex]\int \sqrt{ \tan x} \; dx[/tex]
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
Indefinite integration, also known as antiderivative, is a mathematical process of finding a function that, when differentiated, gives the original function. It is the opposite of differentiation and is used to find the most general form of a function.
Definite integration involves finding the exact value of the area under a curve between two specific points, while indefinite integration involves finding a general function without specific limits.
The main purpose of indefinite integration is to find the most general solution to a differential equation. It is also used in physics, engineering, and other sciences to model and analyze continuous systems and processes.
To solve an indefinite integration problem, you need to follow specific rules and techniques, such as the power rule, substitution, integration by parts, and trigonometric identities. It is also essential to remember the constant of integration, which can change the solution by any arbitrary constant.
Indefinite integration has various real-life applications, such as calculating the position, velocity, and acceleration of an object in motion, determining the growth and decay of populations or radioactive substances, and finding the area and volume of irregular shapes in physics, engineering, economics, and other fields.