`int 1 / sqrt(-x^2-4x) dx` Find or evaluate the integral by completing the square

Thursday, September 6, 2012

$\displaystyle\int\frac{{1}}{\sqrt{{-{{x}}^{{2}}-{4}{x}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

By completing the square and making simple substitution, we will reduce this integral to a table one.

$\displaystyle-{{x}}^{{2}}-{4}{x}=-{\left({{x}}^{{2}}+{4}{x}+{4}\right)}+{4}=-{{\left({x}+{2}\right)}}^{{2}}+{4}={4}-{{\left({x}+{2}\right)}}^{{2}}.$

Now make a substitution $\displaystyle{y}=\frac{{{x}+{2}}}{{2}},$ then $\displaystyle{\left.{d}{y}\right.}=\frac{{\left.{d}{x}}}{{2}},$ $\displaystyle{\left.{d}{x}\right.}={2}{\left.{d}{y}\right.}$ and integral becomes

$\displaystyle\int\frac{{1}}{\sqrt{{{4}-{4}{{y}}^{{2}}}}}\cdot{2}{\left.{d}{y}\right.}=\int\frac{{{\left.{d}{y}\right.}}}{\sqrt{{{1}-{{y}}^{{2}}}}}={\arcsin{{\left({y}\right)}}}+{C}={\arcsin{{\left(\frac{{{x}+{2}}}{{2}}\right)}}}+{C},$

where $\displaystyle{C}$ is any constant.

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Thursday, September 6, 2012

$\displaystyle\int\frac{{1}}{\sqrt{{-{{x}}^{{2}}-{4}{x}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

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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?

Thursday, September 6, 2012

Find or evaluate the integral by completing the square

No comments:

Post a Comment

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

$\displaystyle\int\frac{{1}}{\sqrt{{-{{x}}^{{2}}-{4}{x}}}}{\left.{d}{x}\right.}$ Find or evaluate the integral by completing the square

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