February 18th, 2017, 05:49 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 169 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 07:50 PM. 
February 18th, 2017, 06:13 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 922 
Use the substitution $x=\sin^2(u)$ 
February 18th, 2017, 06:22 AM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 169 Thanks: 2  
February 18th, 2017, 09:44 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 922  
February 18th, 2017, 11:52 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,702 Thanks: 1527 
I assume you mean the $\sqrt{1  x}$ term.

February 18th, 2017, 01:10 PM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 922  
February 18th, 2017, 04:03 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 07:51 PM.  
February 19th, 2017, 01:48 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, 01:52 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, 03:36 AM  #10 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,689 Thanks: 670 Math Focus: Wibbly wobbly timeywimey stuff. 
Again hard to read. And could you please post it right side up? Dan 

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