February 18th, 2017, 06:49 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2  This integral problem kicked my ass. Please help
Integral[ sqrt(x) / sqrt(1x) dx ] Answer: arcsin( sqrt(x) )  sqrt(x) * sqrt(1  x) + C It would be much appreciated if someone points me at the right direction. Last edited by skipjack; February 18th, 2017 at 08:50 PM. 
February 18th, 2017, 07:13 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,197 Thanks: 1152 
Use the substitution $x=\sin^2(u)$ 
February 18th, 2017, 07:22 AM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2  
February 18th, 2017, 10:44 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,197 Thanks: 1152  
February 18th, 2017, 12:52 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,951 Thanks: 1842 
I assume you mean the $\sqrt{1  x}$ term.

February 18th, 2017, 02:10 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,197 Thanks: 1152  
February 18th, 2017, 05:03 PM  #7  
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35  Quote:
Let $\displaystyle \begin{align*} u = 1  \left( x  1 \right) ^{1} \implies \mathrm{d}u = \left( x  1 \right) ^{2} \end{align*}$ and the integral becomes $\displaystyle \begin{align*} \int{ \sqrt{ 1  \left( x  1 \right) ^{1} }\,\left( x  1 \right) ^2\,\left( x  1 \right) ^{2}\,\mathrm{d}x } &= \int{ \sqrt{u}\, \left( 1  u \right) ^{2}\,\mathrm{d}u } \\ &= \int{ \frac{\sqrt{u}}{\left[ 1 + \left( \sqrt{u} \right) ^2 \right] ^2} \,\mathrm{d}u } \\ &= \int{ \frac{2\,\left( \sqrt{u} \right) ^2}{2\,\sqrt{u}\,\left[ 1 + \left( \sqrt{u} \right) ^2 \right] ^2}\,\mathrm{d}u } \end{align*}$ Let $\displaystyle \begin{align*} v = \sqrt{u} \implies \mathrm{d}v = \frac{1}{2\,\sqrt{u}} \end{align*}$ and the integral becomes $\displaystyle \begin{align*} \int{ \frac{2\,\left( \sqrt{u} \right) ^2}{2\,\sqrt{u}\,\left[ 1 + \left( \sqrt{u} \right) ^2 \right] ^2}\,\mathrm{d}u } &= \int{ \frac{2\,v^2}{\left[ 1 + v^2 \right] ^2} \,\mathrm{d}v } \end{align*}$ and now a trigonometric or hyperbolic substitution is a bit more obvious. I would lean towards $\displaystyle \begin{align*} v = \sinh{(t)} \implies \mathrm{d}v = \cosh{(t)}\,\mathrm{d}t \end{align*}$... Last edited by skipjack; February 18th, 2017 at 08:51 PM.  
February 19th, 2017, 02:48 AM  #8 
Newbie Joined: Feb 2017 From: India Posts: 4 Thanks: 0  I tried upto this.... Sent from my Micromax AQ4502 using Tapatalk 
February 19th, 2017, 02:52 AM  #9 
Newbie Joined: Feb 2017 From: India Posts: 4 Thanks: 0 
Plz ignore the last line Sent from my Micromax AQ4502 using Tapatalk 
February 19th, 2017, 04:36 AM  #10 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,914 Thanks: 774 Math Focus: Wibbly wobbly timeywimey stuff. 
Again hard to read. And could you please post it right side up? Dan 

Tags 
ass, integral, kicked, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Integral problem  Jadephilo  Calculus  4  January 19th, 2016 07:54 PM 
Integral problem  motant  Calculus  1  March 12th, 2015 02:05 PM 
Integral problem  azelio  Calculus  2  October 24th, 2010 12:01 PM 
Integral Problem  sostenitore  Calculus  4  December 27th, 2007 10:13 AM 