-w8.3.11 int sqrt{x} dx sqrt{1-x}

[karush]

w8.3.11 nmh{1000}

$\displaystyle I= \int\frac {\sqrt{x}} {\sqrt{1-x}}\ dx=\arcsin\left({\sqrt{x}}\right)-\sqrt{x}\sqrt{1 - x}+C $
Substitutions $x=\sin^2 \left({u}\right)
\quad dx=2\sin\left({u}\right) \cos\left({u}\right) \ du
\quad u=\arcsin\left({\sqrt{x}}\right)$

This evaluates to $I=2\int\sin^2 \left({u}\right) \ du
= {u}-\sin\left({u}\right) \cos\left({u}\right)$
Back substittute u.. $I=\arcsin\left({\sqrt{x}}\right)-\sqrt{x}\sqrt{1 - x}+C $

 

[Prove It]

Whitman 8.3.11

$\displaystyle \int\frac {\sqrt{x}} {\sqrt{1-x}}\ dx
=\arcsin\left({\sqrt{x}}\right)-\sqrt{x}\sqrt{1 - x}+C $
By observation was going to use the trig substitution of
$x=\sin^2 \left({u}\right)$
But this may not merit a trig substitution...

$\displaystyle \begin{align*} \int{ \frac{\sqrt{x}}{\sqrt{1 - x}}\,\mathrm{d}x } &= -2\int{ \sqrt{x}\,\left( -\frac{1}{2\,\sqrt{1 - x}} \right) \,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u = \sqrt{1 - x} \implies \mathrm{d}u = -\frac{1}{2\,\sqrt{1 - x}}\,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} -2\int{ \sqrt{x}\,\left( -\frac{1}{2\,\sqrt{1 - x}} \right) \,\mathrm{d}x } &= -2\int{ \sqrt{1 - u^2}\,\mathrm{d}u } \end{align*}$

You can now continue with $\displaystyle \begin{align*} u = \sin{\left( \theta \right) } \implies \mathrm{d}u = \cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$ to get

$\displaystyle \begin{align*} -2\int{ \sqrt{1 - u^2}\,\mathrm{d}u } &= -2\int{ \sqrt{1 - \sin^2{\left( \theta \right) }}\,\cos{\left( \theta \right) } \,\mathrm{d}\theta } \\ &= -2\int{ \cos^2{ \left( \theta \right) } \,\mathrm{d}\theta } \\ &= -2\int{ \frac{1 + \cos{ \left( 2\,\theta \right) }}{2} \,\mathrm{d}\theta } \\ &= -\int{ \left[ 1 + \cos{ \left( 2\,\theta \right) } \right] \,\mathrm{d}\theta } \\ &= - \left[ \theta + \frac{1}{2}\sin{ \left( 2\,\theta \right) } \right] + C \\ &= - \left[ \theta + \sin{ \left( \theta \right) } \cos{ \left( \theta \right) } \right] + C \\ &= - \left[ \theta + \sin{ \left( \theta \right) } \,\sqrt{ 1 - \sin^2{ \left( \theta \right) } } \right] + C \\ &= - \left[ \arcsin{ \left( u \right) } + u \,\sqrt{ 1 - u^2 } \right] + C \\ &= -\left[ \arcsin{ \left( \sqrt{ 1 - x } \right) } + \sqrt{1 - x} \,\sqrt{ 1 - \left( \sqrt{1 - x} \right) ^2 } \right] + C \\ &= - \left[ \arcsin{ \left( \sqrt{1 - x} \right) } + \sqrt{1 - x} \,\sqrt{ 1 - \left( 1 - x \right) } \right] + C \\ &= - \left[ \arcsin{ \left( \sqrt{1 - x} \right) } + \sqrt{1 - x} \, \sqrt{ x } \right] + C \end{align*}$

 

[karush]

OK that is quite helpful a lot of steps tho.
I didn't think I would get a reply so I pursued the OP

Yours would avoid the
$$2\int\sin^2 \left({u}\right) \ du$$ evaluation

 

[karush]

MHB replies to calculus problems