My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree6Thanks
LinkBack Thread Tools Display Modes
October 6th, 2018, 06:18 PM   #11
Global Moderator
Joined: Dec 2006

Posts: 21,119
Thanks: 2331

You slipped up when doing the initial substitution.

Also, you forgot to include a constant of integration.
skipjack is offline  
October 6th, 2018, 08:31 PM   #12
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,700
Thanks: 2682

Math Focus: Mainly analysis and algebra
Apart from skipjack's comments, it looks good. But if you are going to use a substitution $x=\sec u$, why not do it on the integral $\displaystyle \int \sqrt{x^2-1} \,\mathrm dx$? Admittedly it leads to the mildly challenging $$\int \sec^3 u \, \mathrm du$$
I approached that one via \begin{align}\int \sec{(u)} \, \mathrm du &= \int \cos^2{(u)}\sec^3{(u)}\, \mathrm du \\ &= \int (1-\sin^2{(u)})\sec^3{(u)}\, \mathrm du \\ &= \int \sec^3{(u)}\,\mathrm du - \int \underbrace{\sin{(u)}}_{u} \underbrace{\sin{(u)} \cos^{-3}{(u)}}_{\mathrm dv}\, \mathrm du \end{align}
v8archie is offline  
October 6th, 2018, 09:05 PM   #13
Global Moderator
Joined: Dec 2006

Posts: 21,119
Thanks: 2331

Substituting $x = \cosh u$ is simpler for that (or for integrating $x^2 /\sqrt{x^2 - 1}$) if you know about the hyperbolic functions.
skipjack is offline  
October 21st, 2019, 03:27 PM   #14
Senior Member
Joined: Feb 2016
From: Australia

Posts: 1,847
Thanks: 661

Math Focus: Yet to find out.
Originally Posted by SDK View Post
Anyways, I remembered the following integral which kept me busy for a few hours (or more) in grad school. Enjoy.

$\int_0^{2\pi} \log \Gamma (\frac{x}{2 \pi}) \exp(\cos x) \sin(x + \sin x) \ dx $ where $\Gamma$ is the usual gamma function.
Anyone end up solving this?
Joppy is offline  
October 24th, 2019, 09:19 PM   #15
Senior Member
Joined: Sep 2016
From: USA

Posts: 684
Thanks: 459

Math Focus: Dynamical systems, analytic function theory, numerics
Originally Posted by Joppy View Post
Anyone end up solving this?
I guess not. I might post the solution if I get some time. In any case I think a year is probably plenty long and we can safely skip it. You should feel free to continue the marathon by posting another integral.
SDK is offline  

  My Math Forum > College Math Forum > Calculus

challenging, integral, marathon

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Challenging integral Realistonia Calculus 8 February 27th, 2013 06:42 AM
A Challenging Integral Etyucan Calculus 11 December 7th, 2012 11:55 AM
A challenging integral zaidalyafey Calculus 2 November 17th, 2012 05:56 AM
here comes the 1st challenging integral davedave Calculus 28 November 12th, 2012 03:36 AM
another somewhat challenging contour integral?. galactus Complex Analysis 3 May 15th, 2012 02:39 PM

Copyright © 2019 My Math Forum. All rights reserved.