`int_sqrt(3)^3 1/ (xsqrt(4x^2-9)) dx` Evaluate the definite integral

Sunday, July 8, 2012

$\displaystyle{\int_{\sqrt{{{3}}}}^{{3}}}\frac{{1}}{{{x}\sqrt{{{4}{{x}}^{{2}}-{9}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

Make the substitution $\displaystyle{u}=\sqrt{{{4}{{x}}^{{2}}-{9}}},$ then $\displaystyle{d}{u}=\frac{{{4}{x}}}{\sqrt{{{4}{{x}}^{{2}}-{9}}}}{\left.{d}{x}\right.}.$ Inversely, $\displaystyle{\left.{d}{x}\right.}=\frac{\sqrt{{{4}{{x}}^{{2}}-{9}}}}{{{4}{x}}}{d}{u}=\frac{{u}}{{{4}{x}}}{d}{u}$ and $\displaystyle{4}{{x}}^{{2}}={{u}}^{{2}}+{9}.$ The limits of integration become from $\displaystyle\sqrt{{{3}}}$ to $\displaystyle{3}\sqrt{{{3}}}.$

The indefinite integral becomes

$\displaystyle\int\frac{{u}}{{{4}{u}{{x}}^{{2}}}}{d}{u}=\int\frac{{{d}{u}}}{{{{u}}^{{2}}+{9}}}=\frac{{1}}{{3}}{\arctan{{\left(\frac{{u}}{{3}}\right)}}}+{C},$

where $\displaystyle{C}$ is an arbitrary constant.

Thus the definite integral is $\displaystyle\frac{{1}}{{3}}{\left({\arctan{{\left(\sqrt{{{3}}}\right)}}}-{\arctan{{\left(\frac{{1}}{\sqrt{{{3}}}}\right)}}}\right)}=\frac{{1}}{{3}}{\left(\frac{\pi}{{3}}-\frac{\pi}{{6}}\right)}=\frac{\pi}{{18}}.$

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Sunday, July 8, 2012

$\displaystyle{\int_{\sqrt{{{3}}}}^{{3}}}\frac{{1}}{{{x}\sqrt{{{4}{{x}}^{{2}}-{9}}}}}{\left.{d}{x}\right.}$ Evaluate the definite 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, July 8, 2012

Evaluate the definite 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?

$\displaystyle{\int_{\sqrt{{{3}}}}^{{3}}}\frac{{1}}{{{x}\sqrt{{{4}{{x}}^{{2}}-{9}}}}}{\left.{d}{x}\right.}$ Evaluate the definite integral

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