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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

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