Blog: `int_0^(2pi) t^2 sin(2t) dt` Evaluate the integral

Tuesday, August 30, 2016

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

You need to use the integration by parts for $\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

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

$\displaystyle{d}{v}={\sin{{2}}}{t}\Rightarrow{v}=\frac{{-{\cos{{2}}}{t}}}{{2}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}={{t}}^{{2}}\cdot\frac{{-{\cos{{2}}}{t}}}{{2}}{{\mid}_{{0}}^{{{2}\pi}}}+{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}$

You need to use the integration by parts for $\displaystyle{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}$ such that:

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

$\displaystyle{d}{v}={\cos{{2}}}{t}\Rightarrow{v}=\frac{{{\sin{{2}}}{t}}}{{2}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}={t}\cdot\frac{{{\sin{{2}}}{t}}}{{2}}{{\mid}_{{0}}^{{{2}\pi}}}-{\left(\frac{{1}}{{2}}\right)}{\int_{{0}}^{{{2}\pi}}}{\sin{{2}}}{t}\ldots$

You need to use the integration by parts for $\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

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

$\displaystyle{d}{v}={\sin{{2}}}{t}\Rightarrow{v}=\frac{{-{\cos{{2}}}{t}}}{{2}}$

You need to use the integration by parts for $\displaystyle{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}$ such that:

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

$\displaystyle{d}{v}={\cos{{2}}}{t}\Rightarrow{v}=\frac{{{\sin{{2}}}{t}}}{{2}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{t}\cdot{\cos{{2}}}{t}{\left.{d}{t}\right.}={t}\cdot\frac{{{\sin{{2}}}{t}}}{{2}}{{\left|_{{0}}^{{{2}\pi}}+\frac{{{\cos{{2}}}{t}}}{{4}}\right|}_{{0}}^{{{2}\pi}}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}={{t}}^{{2}}\cdot\frac{{-{\cos{{2}}}{t}}}{{2}}{{\left|_{{0}}^{{{2}\pi}}+{t}\cdot\frac{{{\sin{{2}}}{t}}}{{2}}\right|}_{{0}}^{{{2}\pi}}}+\frac{{{\cos{{2}}}{t}}}{{4}}{{\mid}_{{0}}^{{{2}\pi}}}$

Using the fundamental theorem of calculus yields:

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}={{\left({2}\pi\right)}}^{{2}}\cdot\frac{{-{\cos{{4}}}\pi}}{{2}}+{0}\cdot\frac{{{\cos{{0}}}}}{{2}}+{2}\pi\cdot\frac{{{\sin{{4}}}\pi}}{{2}}-{0}+\frac{{{\cos{{4}}}\pi}}{{4}}-\frac{{{\cos{{0}}}}}{{4}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=-{2}{{\left(\pi\right)}}^{{2}}+\frac{{1}}{{4}}-\frac{{1}}{{4}}$

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=-{2}{{\left(\pi\right)}}^{{2}}$

Hence, evaluating the integral, using integration by parts, yields $\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}\cdot{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}=-{2}{{\left(\pi\right)}}^{{2}}.$

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Tuesday, August 30, 2016

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

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

Tuesday, August 30, 2016

Evaluate the integral

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

$\displaystyle{\int_{{0}}^{{{2}\pi}}}{{t}}^{{2}}{\sin{{\left({2}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

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