Question

Evaluate the integral\(\int {\cos \sqrt x dx} \)

Short answer

We will substitute \(\sqrt x = t\)and then solve further to get the required answer

Step-by-step solution

Given Data

Given = \({\rm I} = \int {\cos \sqrt x dx} \)

substitution

Let, \(\begin{aligned}{l}\sqrt x = t\\\frac{1}{{2\sqrt x }}dx = dt\\\therefore {\rm I} = \int {\cos \left( t \right)2tdt} \end{aligned}\)

Integration by Part

\({\rm I} = 2\int {t\cos tdt} \)

Let,\(\begin{aligned}{l}\mu = t\\d\mu = dt\end{aligned}\) \(\begin{aligned}{l}d\nu = \cos tdt\\\nu = \sin t\end{aligned}\)

\(\begin{aligned}{l}{\rm I} = 2\left( {t\sin t - \int {\sin tdt} } \right)\\ = 2t\sin t + 2\cos t + c\end{aligned}\)

Substitute back\(t = \sqrt x \)

\({\rm I} = 2\left( {\sqrt x \sin \sqrt x + \cos \sqrt x } \right) + c\)

Hence, \(\int {\cos \sqrt x dx} = 2\left( {\sqrt x \sin \sqrt x + \cos \sqrt x } \right) + c\)