Question

Evaluate the integral\(\int\limits_0^1 {t\cosh tdt} \)

Short answer

We will use theintegration by part method to solve this given integral

\(\int {\mu d\nu = \mu \nu } - \int {\nu .} d\mu \)

Step-by-step solution

Given Data

Given = \({\rm I} = \int\limits_0^1 {t\cosh tdt} \)

Let \(\mu = t\) \(d\mu = dt\)

\(\int {d\nu = \int {\cosh tdt} } \) \(\nu = \sinh t\)

use integration by part

\(\begin{aligned}{l}\int\limits_0^1 {t\cosh tdt} = \left( {t\sinh t} \right)_0^1 - \int\limits_0^1 {\sinh tdt} \\ = \sinh - \left( {\cosh t} \right)_0^1\\ = \sinh - \left( {\cosh - 1} \right)\\ = \sinh - \cosh + 1\end{aligned}\)

Hence, \(\int\limits_0^1 {t\cosh tdt} = \sinh \left( 1 \right) - \cosh \left( 1 \right) + 1\)