Blog: `int_0^1 t cosh(t) dt` Evaluate the integral

Take the indefinite integral by parts. It is known (and easy to prove) that $\displaystyle{\cosh{{\left({t}\right)}}}=\frac{{d}}{{\left.{d}{t}}}{\sinh{{\left({t}\right)}}}$ and $\displaystyle{\sinh{{\left({t}\right)}}}=\frac{{d}}{{\left.{d}{t}}}{\cosh{{\left({t}\right)}}}.$

Denote $\displaystyle{u}={t},$ $\displaystyle{v}={\sinh{{\left({t}\right)}}},$ then $\displaystyle{d}{u}={\left.{d}{t}\right.}$ and $\displaystyle{d}{v}={\cosh{{\left({t}\right)}}}{\left.{d}{t}\right.}.$

$\displaystyle\int{t}{\cosh{{\left({t}\right)}}}{\left.{d}{t}\right.}=\int{u}{d}{v}={u}{v}-\int{v}{d}{u}=$

$\displaystyle={t}{\sinh{{\left({t}\right)}}}-\int{\sinh{{\left({t}\right)}}}{\left.{d}{t}\right.}={t}{\sinh{{\left({t}\right)}}}-{\cosh{{\left({t}\right)}}}+{C}.$

Therefore the definite integral is

$\displaystyle{\sinh{{\left({1}\right)}}}-{\cosh{{\left({1}\right)}}}+{1}=\frac{{1}}{{2}}{\left({{e}}^{{1}}-{{e}}^{{-{1}}}-{{e}}^{{1}}-{{e}}^{{-{1}}}\right)}+{1}={1}-\frac{{1}}{{e}}.$