Question

Evaluate the integral\(\int\limits_0^t {{c^s}\sin \left( {t - s} \right)ds} \)

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 =\({\rm I} = \int\limits_0^t {{c^s}\sin \left( {t - s} \right)ds} \)

Let \(\mu = {\left( {\ln x} \right)^2}\) \(d\mu = \frac{{2\ln x}}{x}dx\)

\(\int {d\nu = \int {dx} } \) \(\nu = x\)

Use Integration by Part

\({\rm I} = x{\left( {\ln x} \right)^2} - 2\int {\ln xdx} \)

\(\begin{aligned}{l} = \left( {x{{\left( {\ln x} \right)}^2}} \right)_1^2 - 2\left( {x\ln x - x} \right)_1^2\\ = 2{\left( {\ln 2} \right)^2} - 2\left( {2\ln 2 - 2 + 1} \right)\\ = 2{\left( {\ln 2} \right)^2} - 4\ln 2 + 2\\ = 0.188\end{aligned}\)

Hence, \(\int\limits_1^2 {{{\left( {\ln x} \right)}^2}dx} = 0.188\)