use integration by parts to evaluate ∫ 5x cos...

# Explanation: ## Step1: Recall integration - by - parts formula The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = 5x$ and $\mathrm{d}v=\cos(4x)\mathrm{d}x$. ## Step2: Find $\mathrm{d}u$ and $v$ Differentiate $u = 5x$ with respect to $x$ to get $\mathrm{d}u=5\mathrm{d}x$. Integrate $\mathrm{d}v=\cos(4x)\mathrm{d}x$ with respect to $x$. Let $t = 4x$, then $\mathrm{d}t = 4\mathrm{d}x$ and $v=\int\cos(4x)\mathrm{d}x=\frac{1}{4}\sin(4x)$. ## Step3: Apply the integration - by - parts formula $\int 5x\cos(4x)\mathrm{d}x=5x\times\frac{1}{4}\sin(4x)-\int\frac{1}{4}\sin(4x)\times5\mathrm{d}x$. ## Step4: Evaluate the remaining integral $\int\frac{5}{4}\sin(4x)\mathrm{d}x$. Let $t = 4x$, $\mathrm{d}t = 4\mathrm{d}x$, then $\int\frac{5}{4}\sin(4x)\mathrm{d}x=-\frac{5}{16}\cos(4x)+C$. So $\int 5x\cos(4x)\mathrm{d}x=\frac{5}{4}x\sin(4x)+\frac{5}{16}\cos(4x)+C$. # Answer: $\frac{5}{4}x\sin(4x)+\frac{5}{16}\cos(4x)+C$