MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

Integrate[Sqrt[Tan[x]], {x, 0, 1}]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg30130] Integrate[Sqrt[Tan[x]], {x, 0, 1}]
  • From: seidov at yahoo.com (Zakir F. Seidov)
  • Date: Sun, 29 Jul 2001 21:26:21 -0400 (EDT)
  • Organization: The Math Forum
  • References: <9j8nqi$5qf$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

After some manipulations this integral is:

(2*(Pi + ArcTan[Sqrt[2*Tan[1]]/(1 - Tan[1])])+ 
   Log[(Tan[1] - Sqrt[2*Tan[1]] + 1)/
     (Tan[1] + Sqrt[2*Tan[1]] + 1)])/(2*Sqrt[2]), or

0.7272982493435106435530636987436355542425

Enjoy!

Zakir

%%%%%%%%%%%%%%%%%%%%%%%%%

Subject: [mg30130]      Integrate[Sqrt[Tan[x]], {x, 0, 1}]  --  approx ??
Author:       Tim 9-23 <bendoftimeb at stny.rr.com>
Organization: Steven M. Christensen and Associates, Inc and
MathTensor, Inc.

How do I just get the approx. of this, say 5 decimal places?
Mathematica gives an exact answer which is in terms of multiple
fractions, roots, and complex numbers.

According to my TI-85, it is approx. 0.7273

My email address is anti-spammed.  Remove the 2 B's after hitting
email reply
if you want to email me.

Tim 9-23




  • Prev by Date: Re: Filename as Function Argument"
  • Next by Date: Re: A Penny for Thought
  • Previous by thread: Re: Filename as Function Argument"
  • Next by thread: Integrate[Sqrt[Tan[x]], {x, 0, 1}]