First make a substitution and then use integration$-$by$-$parts to evaluate the integral.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}8cos\sqrt[\phantom{2}]{x}dx$

Step $1$

If we let $w=\sqrt[\phantom{2}]{x},$ then $dw=\frac{1}{\frac{2}{x}}\frac{1}{2\sqrt[\phantom{2}]{x}}dx.$

Step $2$

Now, we have ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}8cos\sqrt[\phantom{2}]{x}dx=8{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(cosw)(\sqrt[2w]{2w})dw.$

Step $3$

This can be rewritten as $16{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$wcoswdw.

In order to evaluate $16{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$wcoswdw, we'll let $u=w$ and $dv=coswdw.$

Then $du=dw$ and $v=$

Step $4$

We have

$16{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$wcoswdw$=16[,-{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}sin(w)dw].$

Step $5$

Now,

$16{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$wcoswdw$=16$wsinw$+16,+C$

Step $6$

Substituting back in $w=\sqrt[\phantom{2}]{x},$ and incorporating the constant of integration, $C,$ we finally have the following.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}8cos\sqrt[\phantom{2}]{x}dx=$