## Calculus: Early Transcendentals 8th Edition

$$\int x\cos 5xdx=\frac{1}{5}x\sin(5x)+\frac{1}{25}\cos(5x)+C$$
$$A=\int x\cos 5xdx$$ 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 $$A=uv-\int vdu$$ $$A=\frac{1}{5}x\sin(5x)-\int\frac{1}{5}\sin(5x)dx$$ $$A=\frac{1}{5}x\sin(5x)-\frac{1}{5}\int\frac{1}{5}\sin(5x)d(5x)$$ $$A=\frac{1}{5}x\sin(5x)-\frac{1}{25}(-\cos(5x)+C)$$ $$A=\frac{1}{5}x\sin(5x)+\frac{1}{25}\cos(5x)+C$$