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sriram

Integral ---srqt(sine x)

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Is it possible to integrate

 

sqrt(sin x)

 

My teacher told, its not? Why?

 

If it is inegrable, how?

 

Thanks!

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What is true is that the intagral of [math] \sqrt{\sin x} [/math] cannot be written in terms of elementary functions. It can be written in terms of the elliptic integral of the second kind

 

[math]E(\phi,m) = \int^{\phi}_{0}(1-m \sin ^{2}\theta)^{1/2}d \theta[/math]

 

with [math] -\pi/2 < \phi < \pi/2 [/math]

 

The integral can be expressed as

 

[math] \sqrt{\sin x}= -2 E(\frac{1}{2}(\frac{\pi}{2} -x),2) [/math]

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What is true is that the intagral of [math] \sqrt{\sin x} [/math] cannot be written in terms of elementary functions. It can be written in terms of the elliptic integral of the second kind

 

[math]E(\phi' date='m) = \int^{\phi}_{0}(1-m \sin ^{2}\theta)^{1/2}d \theta[/math']

 

with [math] -\pi/2 < \phi < \pi/2 [/math]

 

The integral can be expressed as

 

[math] \sqrt{\sin x}= -2 E(\frac{1}{2}(\frac{\pi}{2} -x),2) [/math]

 

Nice idea, But anyway, What will be the final answes after integration of the fn.

 

I need the final answer, soild answer.

 

Thanks!

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Define F(x) to be a function, such that dF/dx = (sin(x))^{1/2}, we can do this since the function on the right hand side is continous (when positive - let's not go into complex stuff) and hence integrable. Now, F is the answer to your question. If you don't like that answer then I can't help you. It is a perfectly good solid final answer. If we normalize so that F(0)=0, it is uniquely determined, and F(t) can be evaluated for any t to any arbitrary degree of precision. I don't see how that cannot be considered 'the final solid answer'. If you disagree, then it is because you think that you must have an answer written down in some 'nice' form that isn't actually necessary.

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Either you have to define a new 'elementary' function, and in fact matt grime did with his 'F' function. This is a perfectly suitable way of reasoning, especially if a certain form of integration becomes very common and is needed many times in different fields of science.

Or you have to resort to numerical methods, which can provide you the value of the integral for any starting point and end point within the validity area of the function sqrt(sin(x)).

If you want an answer in terms of high school standard functions like sqrt(), sin(), cos() etc, then you will probably be disappointed. There is no such answer.

 

Many integrals cannot be expressed in terms of the elementary functions, but this does not mean that they cannot be integrated, so your teacher's answer is not (fully) correct. What he means is that the integral cannot be expressed in terms of standard high school functions.

 

A nice example of extending the set of standard functions is when it comes to integration of functions of the type C*exp(-D*x*x). It is not possible to write integrals of this type of functions in terms of e-powers, sines, cosines, etc. This type of integrals, however, is very common in statistics, but also in certain parts of physiscs. For this reason a new 'standard' function is introduced, the so-called error function, erf(x). With the definition of this function at hand, one now can express integrals of the form exp(Ax²+Bx+C) in terms of the (complex) erf() function, and possibly other highschool standard functions as exp(x).

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I agree with Matt and Woelen, I have given you the "final" answer. I have given you a well defined function that for you can eveluate for any x. It is however, as I stated originally not in terms of elementary functions like polynomials, trigs etc..

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Mathematica will give you the answer that I gave. I tried it!

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umm well, wen you want it itergrated, are you looking for the exact integral, or are u wanting the area under it for a certain purpose? if so, try using simpsons rule with 13 function values :D nice time consumer. you get 2.3816 between 0 and pi, same for 2pi and 3 pi, 4 pi and 5pi so on in the period. o btw if you use 99999998 segments you get 2.3962, i did it by hand :D nah jokes

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What is true is that the intagral of [math] \sqrt{\sin x} [/math] cannot be written in terms of elementary functions. It can be written in terms of the elliptic integral of the second kind

 

[math]E(\phi,m) = \int^{\phi}_{0}(1-m \sin ^{2}\theta)^{1/2}d \theta[/math]

 

with [math] -\pi/2 < \phi < \pi/2 [/math]

 

The integral can be expressed as

 

[math] \sqrt{\sin x}= -2 E(\frac{1}{2}(\frac{\pi}{2} -x),2) [/math]

If I said I understood approximately, say, zero of that, could you help explain? In simple terms that is. Or is this very complex calculs which I have little chance of understanding it?

 

Maybe elliptic integral of the first kind (if that exists, whatever it might be) is easier to learn than the "second kind"?

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Besides the common "special" functions, you know (e.g. sin, cos, exp, log) there are an infinitude of other functions. The functions you know are only "special" in the sense that they are tought at high schools and the underlying concepts are fairly easy to grasp.

 

Unfortunately, integrals of combinations of these functions most of the time cannot be expressed in these function, and when more advanced mathematics is introduced, more "special" functions are introduced for frequently recurring problems.

 

What ajb does is introduce a new special function, the two-parameter function E and having defined this as a special function, the integral now can be expressed again in well-known functions.

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woelen makes a good point here. There are numerous identified and like he said an infinite amount of functions out there. Just some of them have special names and applications.

 

You know sin and cos, but there is also gamma function, error function, Bessel functions, Airy functions, hyperbolic sine and cosine, hypergeometric function, and many many more. These functions should be thought of in exactly the same way as sine and cosine, since the functions are well defined and their values tabulated. Before computers became really common people, would have books with nothing but tables of these functions' values.

 

In general, whenever a mathematician needed a special function to solve problem, they invented it. E was invented, as least partially, to help evaulate integrals of sine. As you keep taking math classes, and solve more and more complicated problems, you will see where these new functions arise just naturally.

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OK, thanks.

 

So is m some constant which can be easily seen? Maybe generally m can be found by looking at the power, m, in [math](sinx)^{m^{-1}}[/math] (thats m^-1). Which would explain why m=2 in this case.

 

Can phi be calculated using some preset rules? I think I'm correctly interpreting this:

[math]\sqrt{\sin x}= -2 E(\frac{1}{2}(\frac{\pi}{2} -x),2)[/math]

(or should it be:

[math]\int \sqrt{\sin x}= -2 E(\frac{1}{2}(\frac{\pi}{2} -x),2)[/math]

???)

as meaning that:

[math]\phi = (\tfrac{1}{2}(\tfrac{\pi}{2} -x))[/math]

is there a general rule for calculating phi?

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I think that's a typo. Just calculate E and re-arrange it to get back to the integral you want. The point is, it can't be represented in terms of "standard" functions; it's not always essential that you know what the said non-standard functions actually are :)

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Question # 1

Show that the area under y=x3 over the interval [0, b] is .

 

Question # 2

 

(a) Evaluate the definite integral using “The First Fundamental Theorem” of calculus.

 

(b) Solve the definite integral by substitution method.

 

 

 

 

Question #3

 

Find the area of the region that is enclosed between the curves

y=x2+4 and x + y=6.

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Question # 1

Show that the area under y=x3 over the interval [0, b] is .

 

Question # 2

 

(a) Evaluate the definite integral using “The First Fundamental Theorem” of calculus.

 

(b) Solve the definite integral by substitution method.

 

 

 

 

Question #3

 

Find the area of the region that is enclosed between the curves

y=x2+4 and x + y=6.

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Q1. [math]\int x^3 dx=\frac {x^4}{4} \therefore \int_0^b x^3 dx = \frac {b^4}{4}[/math]

 

Q2. a- doesnt make sense, b makes no sense either.

 

Q3. I cant be stuffed doing it manually, 15.0323801 approx.

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O sorry, forgot the +C on the first integral.

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