`int (du)/(u sqrt(5 - u^2)) ` Evaluate the integral

$\displaystyle\int\frac{{{d}{u}}}{{{u}\sqrt{{{5}-{{u}}^{{2}}}}}}$

Let

$\displaystyle{u}=\sqrt{{5}}{\sin{{\left(\theta\right)}}}$

$\displaystyle\frac{{{d}{u}}}{{{d}{\left(\theta\right)}}}=\sqrt{{5}}{\cos{{\left(\theta\right)}}}$

$\displaystyle{\left({d}{u}\right)}=\sqrt{{5}}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

$\displaystyle\int\frac{{{d}{u}}}{{{u}\sqrt{{{5}-{{u}}^{{2}}}}}}$

$\displaystyle=\int\frac{{1}}{{\sqrt{{5}}{\sin{{\left(\theta\right)}}}}}\cdot\frac{{\sqrt{{{5}}}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{\sqrt{{{5}-{\left(\sqrt{{5}}{{\sin{{\left(\theta\right)}}}}^{{2}}\right)}}}}$

$\displaystyle=\int\frac{{{\cot{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{\sqrt{{{5}-{5}{{\sin}}^{{2}}\theta}}}$

$\displaystyle=\int\frac{{{\cot{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{\sqrt{{{5}{\left({1}-{{\sin}}^{{2}}\theta\right)}}}}$

$\displaystyle=\int\frac{{{\cot{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{\sqrt{{{5}{{\cos}}^{{2}}\theta}}}$

$\displaystyle=\int\frac{{{\cot{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{{\sqrt{{{5}}}{\cos{{\left(\theta\right)}}}}}$

$\displaystyle=\int\frac{{{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{{\sqrt{{{5}}}{\sin{{\left(\theta\right)}}}{\cos{{\left(\theta\right)}}}}}$

$\displaystyle=\int\frac{{{d}{\left(\theta\right)}}}{{\sqrt{{{5}}}{\sin{{\left(\theta\right)}}}}}$

$\displaystyle=\int\frac{{{\csc{{\left(\theta\right)}}}{d}{\left(\theta\right)}}}{\sqrt{{{5}}}}$

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}\int{\csc{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}{\ln}{\left|\frac{\sqrt{{{5}}}}{{u}}-\frac{\sqrt{{{5}-{{u}}^{{2}}}}}{{u}}\right|}+{C}$

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}\cdot{\ln}{\left|\frac{{\sqrt{{5}}-\sqrt{{{5}-{{u}}^{{2}}}}}}{{u}}\right|}+{C}$

The final answer is

$\displaystyle=\frac{{1}}{\sqrt{{{5}}}}\cdot{\ln}{\left|\frac{{\sqrt{{5}}-\sqrt{{{5}-{{u}}^{{2}}}}}}{{u}}\right|}+{C}$

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Sunday, May 1, 2011

$\displaystyle\int\frac{{{d}{u}}}{{{u}\sqrt{{{5}-{{u}}^{{2}}}}}}$ Evaluate the 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, May 1, 2011

Evaluate the 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?

$\displaystyle\int\frac{{{d}{u}}}{{{u}\sqrt{{{5}-{{u}}^{{2}}}}}}$ Evaluate the integral

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