Integral of 1 - 2*sinx in square root?

In summary, the problem is that the imaginary terms cancel each other, so there is no simpler closed form than Re(-2*i*E(1/4 , 4) ) where E is the incomplete elliptic function of the second kind and Re is the real part. It is the simplest formal expression.
  • #1
coki2000
91
0
Hello PF members,

I want to solve this integral but I cannot find a method.

[itex]\int\sqrt{1 - 2sin(x)}dx[/itex] for 0 < x < ∏/6

Or more generally [itex]\int\sqrt{a - bsin(x)}dx[/itex] for a > b

How can I solve this?

Thanks in advance
 
Last edited:
Physics news on Phys.org
  • #3
There is no imaginary part in the solution.
The imaginary terms cancel each other. Finally, there is no simpler closed form than Re(-2*i*E(1/4 , 4) ) where E is the incomplete elliptic function of the second kind and Re is the real part. It is the simplest formal expression.
http://www.wolframalpha.com/input/?i=Re(-2*i*E(pi/4,4))
Alternatively, you might express the solution on the form of infinite series (rather complicated), or on the form of a real number (approximative solution thanks to direct numerical integration)
 
  • #4
JJacquelin said:
There is no imaginary part in the solution.
The imaginary terms cancel each other. Finally, there is no simpler closed form than Re(-2*i*E(1/4 , 4) ) where E is the incomplete elliptic function of the second kind and Re is the real part. It is the simplest formal expression.
http://www.wolframalpha.com/input/?i=Re(-2*i*E(pi/4,4))
Alternatively, you might express the solution on the form of infinite series (rather complicated), or on the form of a real number (approximative solution thanks to direct numerical integration)

Thank you for your response. But in the definition of the elliptic integral, we have the square of sine in the square root. In my case it is not square of the k*sinx.
 
  • #5
Of course some transformations are necessary to bring back the integral to a standard form of elliptic function.
Be carreful with confusing concurent notations.
 

Attachments

  • EllipticE.JPG
    EllipticE.JPG
    42.2 KB · Views: 1,065
Last edited:
  • #6
Thank you very much for your help :)
 

Related to Integral of 1 - 2*sinx in square root?

1. What is the integral of 1 - 2*sinx in square root?

The integral of 1 - 2*sinx in square root is ∫√(1 - 2*sinx) dx. This integral cannot be solved analytically and requires the use of numerical methods or approximations.

2. How do you solve an integral with a square root?

To solve an integral with a square root, you can use the substitution method, where you substitute u = √x and solve for u, then replace u with √x in the original integral. Another method is to use trigonometric substitutions, where you rewrite the integral in terms of trigonometric functions.

3. Can the integral of 1 - 2*sinx in square root be evaluated using basic integration rules?

No, the integral of 1 - 2*sinx in square root cannot be evaluated using basic integration rules. It requires more advanced techniques such as numerical methods or approximations.

4. Is there a specific range of values for x in which the integral of 1 - 2*sinx in square root can be solved?

No, there is no specific range of values for x in which the integral of 1 - 2*sinx in square root can be solved. The solution method will depend on the specific function and the desired level of accuracy.

5. What are some real-world applications of integrals with square roots?

Integrals with square roots have various applications in physics, engineering, and economics. Some examples include calculating the work done by a variable force, finding the area under a curve, and determining the average value of a function over a certain interval.

Similar threads

Replies
3
Views
1K
  • Calculus
Replies
3
Views
4K
  • Calculus
Replies
6
Views
1K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
8
Views
1K
Replies
3
Views
1K
Replies
3
Views
1K
Back
Top