Blog: `int t sinh(mt) dt` Evaluate the integral

Saturday, November 13, 2010

$\displaystyle\int{t}{\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

To help you solve this, we consider the the integration by parts:

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

Let $\displaystyle{u}={t}$ and $\displaystyle{d}{v}={\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}.$

based from $\displaystyle\int{t}\cdot{\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$ for $\displaystyle\int{u}\cdot{d}{v}$

In this integral, the "m" will be treated as constant since it is integrated with respect to "t".

From $\displaystyle{u}={t}$ , then $\displaystyle{d}{u}={\left.{d}{t}\right.}$

From $\displaystyle{d}{v}={\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$ , then int dv = v

In $\displaystyle\int{\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$ , let $\displaystyle{w}={m}{t}$ then $\displaystyle{d}{w}={m}{\left.{d}{t}\right.}$ or $\displaystyle{\left.{d}{t}\right.}=\frac{{{d}{w}}}{{m}}$

Substitute $\displaystyle{w}={m}{t}$ and $\displaystyle{\left.{d}{t}\right.}=\frac{{{d}{w}}}{{m}}$

$\displaystyle\int{\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$ = $\displaystyle\int\frac{{{\sinh{{\left({w}\right)}}}{d}{w}}}{{m}}$

= $\displaystyle{\left(\frac{{1}}{{m}}\right)}\int{\sinh{{\left({w}\right)}}}{d}{w}$

based from c is constant in $\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}+{C}$

$\displaystyle{\left(\frac{{1}}{{w}}\right)}\int{\sinh{{\left({w}\right)}}}{d}{w}={\left(\frac{{1}}{{w}}\right)}{\cosh{{\left({w}\right)}}}+{C}$

Substitute $\displaystyle{w}={m}{t}$ , it becomes $\displaystyle{v}=\frac{{1}}{{{m}}}{\cosh{{\left({m}{t}\right)}}}+{C}$

Then:

$\displaystyle{u}={t}$

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

$\displaystyle{d}{v}={\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle{v}=\frac{{1}}{{{m}}}{\cosh{{\left({m}{t}\right)}}}$

Plug into the integration by parts: $\displaystyle\int{u}\cdot{d}{v}={u}{v}-\int{v}\cdot{d}{u}$

$\displaystyle\int{t}\cdot{\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}={t}\cdot\frac{{1}}{{{m}}}{\cosh{{\left({m}{t}\right)}}}-\int\frac{{1}}{{m}}{\cosh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle=\frac{{t}}{{m}}{\cosh{{\left({m}{t}\right)}}}-\frac{{1}}{\min}{t}{\cosh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$

$\displaystyle=\frac{{t}}{{m}}{\cosh{{\left({m}{t}\right)}}}-\frac{{1}}{{m}}\cdot\frac{{1}}{{m}}{\sinh{{\left({m}{t}\right)}}}+{C}$

= $\displaystyle\frac{{t}}{{m}}{\cosh{{\left({m}{t}\right)}}}-\frac{{1}}{{{m}}^{{2}}}{\sinh{{\left({m}{t}\right)}}}+{C}$

= $\displaystyle\frac{{{m}{t}{\cosh{{\left({m}{t}\right)}}}-{\sinh{{\left({m}{t}\right)}}}}}{{{m}}^{{2}}}+{C}$

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Saturday, November 13, 2010

$\displaystyle\int{t}{\sinh{{\left({m}{t}\right)}}}{\left.{d}{t}\right.}$ Evaluate the integral

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