question evaluate the integral below. ∫₀^(π/2) -2x cos(5x) dx be sure to place the argument of any…

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int_{a}^{b}u\mathrm{d}v=uv|{a}^{b}-\int{a}^{b}v\mathrm{d}u$. Let $u = - 2x$ and $\mathrm{d}v=\cos(5x)\mathrm{d}x$. Then $\mathrm{d}u=-2\mathrm{d}x$ and $v=\frac{1}{5}\sin(5x)$. [ \begin{align*} \int_{0}^{\frac{\pi}{2}}-2x\cos(5x)\mathrm{d}x&=\left[-2x\cdot\frac{1}{5}\sin(5x)\right]{0}^{\frac{\pi}{2}}-\int{0}^{\frac{\pi}{2}}\frac{1}{5}\sin(5x)\cdot(- 2)\mathrm{d}x\ \end{align*} ]

Step2: Evaluate the first - term $\left[-2x\cdot\frac{1}{5}\sin(5x)\right]_{0}^{\frac{\pi}{2}}$

When $x = \frac{\pi}{2}$, $-2x\cdot\frac{1}{5}\sin(5x)=-\frac{2\pi}{2}\cdot\frac{1}{5}\sin\left(\frac{5\pi}{2}\right)=-\frac{\pi}{5}\cdot1 =-\frac{\pi}{5}$. When $x = 0$, $-2x\cdot\frac{1}{5}\sin(5x)=0$. So, $\left[-2x\cdot\frac{1}{5}\sin(5x)\right]_{0}^{\frac{\pi}{2}}=-\frac{\pi}{5}-0 =-\frac{\pi}{5}$.

Step3: Evaluate the second - term $\int_{0}^{\frac{\pi}{2}}\frac{2}{5}\sin(5x)\mathrm{d}x$

Let $t = 5x$, then $\mathrm{d}t = 5\mathrm{d}x$ and $\int\frac{2}{5}\sin(5x)\mathrm{d}x=\frac{2}{25}\int\sin(t)\mathrm{d}t=-\frac{2}{25}\cos(t)+C=-\frac{2}{25}\cos(5x)+C$. [ \begin{align*} \int_{0}^{\frac{\pi}{2}}\frac{2}{5}\sin(5x)\mathrm{d}x&=\left[-\frac{2}{25}\cos(5x)\right]_{0}^{\frac{\pi}{2}}\ &=-\frac{2}{25}\cos\left(\frac{5\pi}{2}\right)+\frac{2}{25}\cos(0)\ &=0 + \frac{2}{25}\ &=\frac{2}{25} \end{align*} ]

Step4: Combine the results

[ \begin{align*} \int_{0}^{\frac{\pi}{2}}-2x\cos(5x)\mathrm{d}x&=-\frac{\pi}{5}-\frac{2}{25}\ &=-\frac{5\pi + 2}{25} \end{align*} ]

Answer:

$-\frac{5\pi + 2}{25}$