Evaluate the integral.

$9$xcos$(4$x$)$dx$9$xcos$(4$x$)$dx$$

To use the integration$-$by$-$parts formula $$u$$dv$=$uv$$v$$du$$u$$dv$=$uv$-$v$$du$,$ we must choose one part of $9$xcos$(4$x$)$dx$9$xcos$(4$x$)$dx to be uu$,$ with the rest becoming dvdv$.$

Since the goal is to produce a simpler integral, we will choose u$=9$xu$=9$x$.$ This means that

$$dv$=$dv$=$ dxdx$.$

Now, since u$=9$xu$=9$x$,$ then du$=$du$=$ dxdx$.$

With our choice that dv$=$cos$(4$x$)$dxdv$=$cos$(4$x$)$dx$,$ then v$=$cos$(4$x$)$dxv$=$cos$(4$x$)$dx$.$ This can be calculated using integration by substitution.

In this case $($ignoring the constant of integration$)$ we get v$=$v$=.$

Now, the integration$-$by$-$parts formula $$u$$dv$=$uv$$v$$du$$u$$dv$=$uv$-$v$$du gives us

$$u$$dv$=$uv$$v$$du$$u$$dv$=$uv$-$v$$du$$

$=94$xsin$(4$x$)94$sin$(=94$xsin$(4$x$)-94$sin$($ x$)$dxx$)$dx$.$

We must use substitution to do this second integral.

We can use the substitution t$=4$xt$=4$x$,$ which will give dx$=$dx$=$ dtdt$.$

Ignoring the constant of integration, we have

$$sin$(4$x$)$dx$=$sin$(4$x$)$dx$=.$

Combining our results and including the constant of integration, CC$,$ we finally get

$9$xcos$(4$x$)$dx$=$