`int 1 / (sqrt(x)sqrt(1-x)) dx` Find the indefinite integral

Sunday, October 17, 2010

$\displaystyle\int\frac{{1}}{{\sqrt{{{x}}}\sqrt{{{1}-{x}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Recall that $\displaystyle\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({x}\right)}+{C}$ where:

f(x) as the integrand function

F(x) as the antiderivative of f(x)

C as the constant of integration..

For the given problem, the integral: $\displaystyle\int\frac{{1}}{{\sqrt{{{x}}}\sqrt{{{1}-{x}}}}}{\left.{d}{x}\right.}$

does not yet resemble any formula from table of integrals.

To evaluate this, we have to apply u-substitution by letting:

$\displaystyle{u}=\sqrt{{{x}}}$

Square both sides: $\displaystyle{{\left({u}\right)}}^{{2}}={{\left(\sqrt{{{x}}}\right)}}^{{2}}$ , we get: $\displaystyle{{u}}^{{2}}={x}$

Then plug-in $\displaystyle{{u}}^{{2}}={x}$ in $\displaystyle\sqrt{{{1}-{x}}}$ :

$\displaystyle\sqrt{{{1}-{x}}}=\sqrt{{{1}-{{u}}^{{2}}}}$ .

Apply implicit differentiation on $\displaystyle{{u}}^{{2}}={x}$ , we get: $\displaystyle{2}{u}{d}{u}={\left.{d}{x}\right.}$ .

Plug-in $\displaystyle\sqrt{{{x}}}={u}$ , $\displaystyle\sqrt{{{1}-{x}}}=\sqrt{{{1}-{{u}}^{{2}}}}$ , and $\displaystyle{\left.{d}{x}\right.}={2}{u}{d}{u}$ , we get:

$\displaystyle\int\frac{{1}}{{\sqrt{{{x}}}\sqrt{{{1}-{x}}}}}{\left.{d}{x}\right.}=\int\frac{{1}}{{{u}\cdot\sqrt{{{1}-{{u}}^{{2}}}}}}\cdot{\left({2}{u}{d}{u}\right)}$

$\displaystyle=\int\frac{{{2}{u}{d}{u}}}{{{u}\sqrt{{{1}-{{u}}^{{2}}}}}}$

Cancel out common factor u:

$\displaystyle\int\frac{{1}}{{{u}\sqrt{{{1}-{{u}}^{{2}}}}}}\cdot{\left({2}{u}{d}{u}\right)}=\int\frac{{{2}{d}{u}}}{\sqrt{{{1}-{{u}}^{{2}}}}}$

Apply the basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ :

$\displaystyle\int\frac{{{2}{d}{u}}}{{\sqrt{{{1}-{{u}}^{{2}}}}}}={2}\int\frac{{{d}{u}}}{\sqrt{{{1}-{{u}}^{{2}}}}}$

The integral part resembles the basic integration formula for inverse sine function:

$\displaystyle\int\frac{{{d}{u}}}{\sqrt{{{\left({{a}}^{{2}}-{{u}}^{{2}}\right)}}}}={\arcsin{{\left(\frac{{u}}{{a}}\right)}}}+{C}$

Then,

$\displaystyle{2}\int\frac{{{d}{u}}}{\sqrt{{{1}-{{u}}^{{2}}}}}={2}{\arcsin{{\left(\frac{{u}}{{1}}\right)}}}+{C}$

$\displaystyle={2}{\arcsin{{\left({u}\right)}}}+{C}$

Express it in terms of x by plug-in $\displaystyle{u}=\sqrt{{{x}}}$ for the final answer :

$\displaystyle\int\frac{{1}}{{\sqrt{{{x}}}\sqrt{{{1}-{x}}}}}{\left.{d}{x}\right.}={2}{\arcsin{{\left(\sqrt{{{x}}}\right)}}}+{C}$

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Sunday, October 17, 2010

$\displaystyle\int\frac{{1}}{{\sqrt{{{x}}}\sqrt{{{1}-{x}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Sunday, October 17, 2010

Find the indefinite integral

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle\int\frac{{1}}{{\sqrt{{{x}}}\sqrt{{{1}-{x}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?