evaluate the following integral. ∫₀⁸π t² sin(7t) dt not quite. try again.

Explanation:

Step1: Apply integration - by - parts formula $\int u;dv=uv-\int v;du$

Let $u = t^{2}$, $dv=\sin(7t)dt$. Then $du = 2t;dt$, $v=-\frac{1}{7}\cos(7t)$. [ \begin{align*} \int t^{2}\sin(7t)dt&=-\frac{1}{7}t^{2}\cos(7t)+\frac{2}{7}\int t\cos(7t)dt \end{align*} ]

Step2: Apply integration - by - parts again on $\int t\cos(7t)dt$

Let $u = t$, $dv=\cos(7t)dt$. Then $du = dt$, $v=\frac{1}{7}\sin(7t)$. [ \begin{align*} \int t\cos(7t)dt&=\frac{1}{7}t\sin(7t)-\frac{1}{7}\int\sin(7t)dt\ &=\frac{1}{7}t\sin(7t)+\frac{1}{49}\cos(7t)+C \end{align*} ]

Step3: Combine the results

[ \begin{align*} \int t^{2}\sin(7t)dt&=-\frac{1}{7}t^{2}\cos(7t)+\frac{2}{7}\left(\frac{1}{7}t\sin(7t)+\frac{1}{49}\cos(7t)\right)+C\ &=-\frac{1}{7}t^{2}\cos(7t)+\frac{2}{49}t\sin(7t)+\frac{2}{343}\cos(7t)+C \end{align*} ]

Step4: Evaluate the definite integral

[ \begin{align*} &\left[-\frac{1}{7}t^{2}\cos(7t)+\frac{2}{49}t\sin(7t)+\frac{2}{343}\cos(7t)\right]_0^{8\pi}\ &=-\frac{1}{7}(8\pi)^{2}\cos(56\pi)+\frac{2}{49}(8\pi)\sin(56\pi)+\frac{2}{343}\cos(56\pi)-\left(-\frac{1}{7}(0)^{2}\cos(0)+\frac{2}{49}(0)\sin(0)+\frac{2}{343}\cos(0)\right)\ &=-\frac{64\pi^{2}}{7}(1)+\frac{16\pi}{49}(0)+\frac{2}{343}(1)-\left(0 + 0+\frac{2}{343}(1)\right)\ &=-\frac{64\pi^{2}}{7} \end{align*} ]

Answer:

$-\frac{64\pi^{2}}{7}$