7.1 Integration By Parts/3

${\begin{aligned}\int \,x\cos 5xdx\\[2ex]u=x\qquad dv=\cos 5xdx\\[2ex]du=dx\qquad v={\frac {1}{5}}\sin 5x\\[2ex]\int \,x\cos 5xdx&=x\cdot {\frac {1}{5}}\sin 5x-\int \\[2ex]&=\int \,{\frac {1}{5}}\sin 5x\cdot \,dx\\[2ex]&={\frac {1}{5}}\int \,\sin 5xdx\\[2ex]&={\frac {1}{5}}\cdot {\frac {1}{5}}(-\cos 5x)\\[2ex]&=-{\frac {1}{25}}\cos 5x+C\\[2ex]&={\frac {1}{5}}x\sin 5x+{\frac {1}{25}}\cos 5x+C\end{aligned}}$