Answer
$$\int x\cos 5xdx=\frac{1}{5}x\sin(5x)+\frac{1}{25}\cos(5x)+C$$
Work Step by Step
$$A=\int x\cos 5xdx$$
We would choose $u=x$ and $dv=\cos 5xdx$
We could easily see that $du=dx$.
For $dv=\cos 5xdx$, we analyze as follows $$\int \cos 5xdx$$
Let $u=5x$, then $du=5dx$, so $dx=\frac{1}{5}du$ $$\int\cos 5xdx=\frac{1}{5}\int\cos udu=\frac{1}{5}\sin u+C=\frac{1}{5}\sin(5x)+C$$
Therefore, for $dv=\cos 5xdx$, $v=\frac{1}{5}\sin(5x)$
Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=\frac{1}{5}x\sin(5x)-\int\frac{1}{5}\sin(5x)dx$$ $$A=\frac{1}{5}x\sin(5x)-\frac{1}{5}\int\frac{1}{5}\sin(5x)d(5x)$$ $$A=\frac{1}{5}x\sin(5x)-\frac{1}{25}(-\cos(5x)+C)$$ $$A=\frac{1}{5}x\sin(5x)+\frac{1}{25}\cos(5x)+C$$
