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 February 18th, 2017, 05:49 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 This integral problem kicked my ass. Please help Integral[ sqrt(x) / sqrt(1-x) 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: 2,369 Thanks: 1273 Use the substitution $x=\sin^2(u)$ Thanks from topsquark February 18th, 2017, 06:22 AM   #3
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Brilliant!! May I ask what clue gives you such an elegant substitution?

Quote:
 Originally Posted by romsek Use the substitution $x=\sin^2(u)$ February 18th, 2017, 09:44 AM   #4
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Quote:
 Originally Posted by zollen Brilliant!! May I ask what clue gives you such an elegant substitution?
the $\sqrt{1-x^2}$ term February 18th, 2017, 11:52 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 I assume you mean the $\sqrt{1 - x}$ term. February 18th, 2017, 01:10 PM   #6
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Quote:
 Originally Posted by skipjack I assume you mean the $\sqrt{1 - x}$ term.
yes, sorry.

it becomes $\sqrt{1-\sin^2(u)}=\pm \cos(u)$ February 18th, 2017, 04:03 PM   #7
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Quote:
 Originally Posted by zollen Integral[ sqrt(x) / sqrt(1-x) dx ] Answer: arcsin( sqrt(x) ) - sqrt(x) * sqrt(1 - x) + C It would be much appreciated if someone points me at the right direction.
\displaystyle \begin{align*} \int{ \frac{\sqrt{x}}{\sqrt{1 - x}}\,\mathrm{d}x } &= \int{ \sqrt{ \frac{x}{1 - x} }\,\mathrm{d}x } \\ &= \int{ \sqrt{ -\left( \frac{x}{x - 1} \right) }\,\mathrm{d}x } \\ &= \int{ \sqrt{-\left( 1 + \frac{1}{x - 1 } \right) }\,\mathrm{d}x } \\ &= \int{ \sqrt{ -1 - \left( x - 1 \right) ^{-1} } \,\mathrm{d}x } \\ &= \int{ \sqrt{ -1 - \left( x - 1 \right) ^{-1} }\,\left( x - 1 \right) ^2\,\left( x - 1 \right) ^{-2} \,\mathrm{d}x } \end{align*}

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: 2,073 Thanks: 843 Math Focus: Wibbly wobbly timey-wimey stuff. Again hard to read. And could you please post it right side up? -Dan Tags ass, integral, kicked, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jadephilo Calculus 4 January 19th, 2016 06:54 PM motant Calculus 1 March 12th, 2015 01:05 PM azelio Calculus 2 October 24th, 2010 11:01 AM sostenitore Calculus 4 December 27th, 2007 09:13 AM

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