`int x sqrt(1 - x^4) dx` Evaluate the integral

Monday, July 29, 2013

$\displaystyle\int{x}\sqrt{{{1}-{{x}}^{{4}}}}{\left.{d}{x}\right.}$ Evaluate the integral

You need to solve the integral using substitution method, such that:

$\displaystyle{{x}}^{{2}}={t}\Rightarrow{2}{x}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\Rightarrow{x}{\left.{d}{x}\right.}=\frac{{{\left.{d}{t}\right.}}}{{2}}$

Replacing x by t, yields:

$\displaystyle\int{x}\cdot\sqrt{{{1}-{{x}}^{{4}}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}\int\sqrt{{{1}-{{t}}^{{2}}}}\cdot{\left.{d}{t}\right.}$

You need to use the next trigonometric substitution, such that:

$\displaystyle{t}={\sin{{u}}}\Rightarrow{\left.{d}{t}\right.}={\cos{{u}}}{d}{u}$

$\displaystyle{u}={\arcsin{{t}}}={{\arcsin{{x}}}}^{{2}}$

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

You need to use the fundamental trigonometric formula $\displaystyle{1}-{{\sin}}^{{2}}{u}={{\cos}}^{{2}}{u}$

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

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

Using the half angle formula, yields:

$\displaystyle{{\cos}}^{{2}}{u}=\frac{{{1}+{\cos{{2}}}{u}}}{{2}}$

$\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot\int{\left({{\cos}}^{{2}}{u}\right)}{d}{u}={\left(\frac{{1}}{{4}}\right)}\cdot\int{\left({1}+{\cos{{2}}}{u}\right)}{d}{u}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\cdot\int{\left({1}+{\cos{{2}}}{u}\right)}{d}{u}={\left(\frac{{1}}{{4}}\right)}\cdot\int{d}{u}+{\left(\frac{{1}}{{4}}\right)}\cdot\int{\left({\cos{{2}}}{u}\right)}{d}{u}$

$\displaystyle{\left(\frac{{1}}{{4}}\right)}\cdot\int{\left({1}+{\cos{{2}}}{u}\right)}{d}{u}={\left(\frac{{1}}{{4}}\right)}\cdot{\left({u}+\frac{{{\sin{{\left({2}{u}\right)}}}}}{{2}}\right)}$

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

Hence, evaluating the indefinite integral yields $\displaystyle\int{x}\cdot\sqrt{{{1}-{{x}}^{{4}}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{4}}\right)}\cdot{\left({\arcsin{{\left({{x}}^{{2}}\right)}}}+\frac{{{\sin{{\left({2}{\arcsin{{\left({{x}}^{{2}}\right)}}}\right)}}}}}{{2}}\right)}+{c}.$

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Monday, July 29, 2013

$\displaystyle\int{x}\sqrt{{{1}-{{x}}^{{4}}}}{\left.{d}{x}\right.}$ 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?

Monday, July 29, 2013

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?

$\displaystyle\int{x}\sqrt{{{1}-{{x}}^{{4}}}}{\left.{d}{x}\right.}$ Evaluate the integral

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