Solve Improper Integral of 1/(1+(tanx)^sqrt(pi)) from 0 to pi/2

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Homework Help Overview

The discussion revolves around evaluating an improper integral of the form ∫(1/(1+(tan(x))^√π)) dx from 0 to π/2. Participants express varying levels of difficulty with the integration techniques required for this problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss different methods attempted, including substitution and the use of software like Mathematica. Questions arise about the nature of the antiderivative and whether it can be expressed in elementary terms. There is also a focus on the correct notation for the integral.

Discussion Status

Some participants have provided guidance on using LaTeX for mathematical expressions, while others are exploring the implications of using hypergeometric functions. The conversation reflects a mix of attempts to clarify notation and the mathematical challenges posed by the integral.

Contextual Notes

There are mentions of potential confusion regarding the notation and spelling of "pi," as well as the context in which the integral was encountered, suggesting that the problem may be part of a larger study or application.

SeReNiTy
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Hi i am completely stuck on a intergration problem, i have tried substition, and every technique i can think off but i cannot solve it. It is a improper intergral and the answer is pie/4

The question is:

find the intergral with limits upper pie/2 and lower 0

1/(1+(tanx)^sqaureroot pie).dx

P.S: how do i type equations liek the way some ppl do in their threads
 
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[tex]\int\frac{1}{1+tan^{\sqrt{e\pi}}(x)}{dx}[/tex]

Is that the right integral? Clicking the graphic will show you the code to type it.
 
Okay.Here's the antiderivative

[tex]\int \frac{dx}{1+\tan^{\sqrt{\pi}} x}[/tex]


(v.attachement).

Now,u can see that,both in the lower limit & in the upper one,u need to compute the limits in ivolving the [tex]_{2}F_{1}[/tex] of an adorable argument & tangent of "x" to a weird power.

Good luck with those limits... :smile:

Daniel.
 

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and by the way, [itex]\pi[/itex] is spelled "pi." :)
 
I like "pie"...:-pYumm-Yumm :!) (Please work on that Latex code,will u?People are confused because of your weird notations).

Daniel.
 
There's also only 1 r in both "integration" and "integral".
 
And it's "...think of"...:-p And "like" (sic!)...


Daniel.


P.S.Another piece of advice:use the spell checker.
 
Last edited:
Its "Another PIECE OF advice" :-D
 
  • #10
Thanks so much guys, wow you guys run a real good service. Oh, and thanks for the correction, of course how stupid was i to write "pie", indeed it is "pi".
 
  • #11
whozum said:
[tex]\int\frac{1}{1+tan^{\sqrt{e\pi}}(x)}{dx}[/tex]

Is that the right integral? Clicking the graphic will show you the code to type it.

No, this is the integral [tex]\int \frac{dx}{1+\tan^{\sqrt{\pi}} x}[/tex]
Oh, and to the dexter, could you clarify what steps you took reach the antiderrivative? Like substitution or interation by parts?
 
  • #12
He got mathematica to do it. I find it extremely unlikely that it is possible to express the antiderivative in terms of elementary functions. In what context did this integral appear?
 
  • #13
Data said:
He got mathematica to do it. I find it extremely unlikely that it is possible to express the antiderivative in terms of elementary functions. In what context did this integral appear?

Oh, so he used a computer program or something? So this integral cannot be expressed with elementary functions?
 
  • #14
Well,it cannot.If u decide to include hypergeometric functions among elementary functions,then u might say it is,else,not.

Now,if u plot the functions

[tex]\frac{1}{1+\tan^{\sqrt{\pi}}x}[/tex] & [tex]\frac{1}{1+\tan^{2}x}[/tex] on the interval [tex][0,\frac{\pi}{2})[/tex]

,then,by evaluating the integral of the second function on the same interval (which is very,very easy),u might get an approximation to the first integral.If u have a software that could plot them together in the same picture/graph,it would be perfect.

Daniel.
 

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