Blog: `int t^2 sin(beta t ) dt ` Evaluate the integral

Tuesday, November 5, 2013

$\displaystyle\int{{t}}^{{2}}{\sin{{\left(\beta{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

You need to use the substitution $\displaystyle\beta\cdot{t}={u}$ , such that:

$\displaystyle\beta\cdot{t}={u}\Rightarrow\beta{\left.{d}{t}\right.}={d}{u}$

Replacing the variable, yields:

$\displaystyle\int{{t}}^{{2}}\cdot{\sin{{\left(\beta\cdot{t}\right)}}}{\left.{d}{t}\right.}=\frac{{1}}{{{\beta}^{{3}}}}\int{{u}}^{{2}}\cdot{\sin{{u}}}{d}{u}$

You need to use the integration by parts such that:

$\displaystyle\int{f{{d}}}{g{=}}{f{{g{-}}}}\int{g{{d}}}{f{}}$

$\displaystyle{f{=}}{{u}}^{{2}}\Rightarrow{d}{f{=}}{2}{u}{d}{u}$

$\displaystyle{d}{g{=}}{\sin{{u}}}\Rightarrow{g{=}}-{\cos{{u}}}$

$\displaystyle\frac{{1}}{{{\beta}^{{3}}}}\int{{u}}^{{2}}\cdot{\sin{{u}}}{d}{u}=\frac{{1}}{{{\beta}^{{3}}}}{\left(-{{u}}^{{2}}\cdot{\cos{{u}}}+{2}\int{u}\cdot{\cos{{u}}}{d}{u}\right)}$

You need to use the integration by parts...

You need to use the substitution $\displaystyle\beta\cdot{t}={u}$ , such that:

$\displaystyle\beta\cdot{t}={u}\Rightarrow\beta{\left.{d}{t}\right.}={d}{u}$

Replacing the variable, yields:

$\displaystyle\int{{t}}^{{2}}\cdot{\sin{{\left(\beta\cdot{t}\right)}}}{\left.{d}{t}\right.}=\frac{{1}}{{{\beta}^{{3}}}}\int{{u}}^{{2}}\cdot{\sin{{u}}}{d}{u}$

You need to use the integration by parts such that:

$\displaystyle\int{f{{d}}}{g{=}}{f{{g{-}}}}\int{g{{d}}}{f{}}$

$\displaystyle{f{=}}{{u}}^{{2}}\Rightarrow{d}{f{=}}{2}{u}{d}{u}$

$\displaystyle{d}{g{=}}{\sin{{u}}}\Rightarrow{g{=}}-{\cos{{u}}}$

$\displaystyle\frac{{1}}{{{\beta}^{{3}}}}\int{{u}}^{{2}}\cdot{\sin{{u}}}{d}{u}=\frac{{1}}{{{\beta}^{{3}}}}{\left(-{{u}}^{{2}}\cdot{\cos{{u}}}+{2}\int{u}\cdot{\cos{{u}}}{d}{u}\right)}$

You need to use the integration by parts again, such that:

$\displaystyle{2}\int{u}\cdot{\cos{{u}}}{d}{u}={2}{u}\cdot{\sin{{u}}}-{2}\int{\sin{{u}}}{d}{u}$

$\displaystyle{f{=}}{u}\Rightarrow{d}{f{=}}{d}{u}$

$\displaystyle{d}{g{=}}{\cos{{u}}}\Rightarrow{g{=}}{\sin{{u}}}$

$\displaystyle{2}\int{u}\cdot{\cos{{u}}}{d}{u}={2}{u}\cdot{\sin{{u}}}+{2}{\cos{{u}}}+{c}$

$\displaystyle\frac{{1}}{{{\beta}^{{3}}}}\int{{u}}^{{2}}\cdot{\sin{{u}}}{d}{u}=\frac{{1}}{{{\beta}^{{3}}}}{\left(-{{u}}^{{2}}\cdot{\cos{{u}}}+{2}{u}\cdot{\sin{{u}}}+{2}{\cos{{u}}}\right)}+{c}$

Replacing back the variable, yields:

$\displaystyle\int{{t}}^{{2}}\cdot{\sin{{\left(\beta\cdot{t}\right)}}}{\left.{d}{t}\right.}=\frac{{1}}{{{\beta}^{{3}}}}{\left(-{{\left(\beta\cdot{t}\right)}}^{{2}}\cdot{\cos{{\left(\beta\cdot{t}\right)}}}+{2}{\left(\beta\cdot{t}\right)}\cdot{\sin{{\left(\beta\cdot{t}\right)}}}+{2}{\cos{{\left(\beta\cdot{t}\right)}}}\right)}+{c}$

Hence, evaluating the integral, using substitution, then integration by parts, yields $\displaystyle\int{{t}}^{{2}}\cdot{\sin{{\left(\beta\cdot{t}\right)}}}{\left.{d}{t}\right.}=\frac{{1}}{{{\beta}^{{3}}}}{\left(-{{\left(\beta\cdot{t}\right)}}^{{2}}\cdot{\cos{{\left(\beta\cdot{t}\right)}}}+{2}{\left(\beta\cdot{t}\right)}\cdot{\sin{{\left(\beta\cdot{t}\right)}}}+{2}{\cos{{\left(\beta\cdot{t}\right)}}}\right)}+{c}$

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Tuesday, November 5, 2013

$\displaystyle\int{{t}}^{{2}}{\sin{{\left(\beta{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Tuesday, November 5, 2013

Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\int{{t}}^{{2}}{\sin{{\left(\beta{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?