evaluate the following integral. ∫₀⁴π t² sin(8t) dt

Explanation:

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

Let $u = t^{2}$, $dv=\sin(8t)dt$. Then $du = 2t;dt$, $v=-\frac{1}{8}\cos(8t)$. [ \begin{align*} \int t^{2}\sin(8t)dt&=-\frac{1}{8}t^{2}\cos(8t)+\frac{1}{4}\int t\cos(8t)dt \end{align*} ]

Step2: Apply integration - by - parts again on $\int t\cos(8t)dt$

Let $u = t$, $dv=\cos(8t)dt$. Then $du = dt$, $v=\frac{1}{8}\sin(8t)$. [ \begin{align*} \int t\cos(8t)dt&=\frac{1}{8}t\sin(8t)-\frac{1}{8}\int\sin(8t)dt\ &=\frac{1}{8}t\sin(8t)+\frac{1}{64}\cos(8t)+C \end{align*} ]

Step3: Combine the results

[ \begin{align*} \int t^{2}\sin(8t)dt&=-\frac{1}{8}t^{2}\cos(8t)+\frac{1}{4}\left(\frac{1}{8}t\sin(8t)+\frac{1}{64}\cos(8t)\right)+C\ &=-\frac{1}{8}t^{2}\cos(8t)+\frac{1}{32}t\sin(8t)+\frac{1}{256}\cos(8t)+C \end{align*} ]

Step4: Evaluate the definite integral

[ \begin{align*} &\left[-\frac{1}{8}t^{2}\cos(8t)+\frac{1}{32}t\sin(8t)+\frac{1}{256}\cos(8t)\right]_0^{4\pi}\ &=-\frac{1}{8}(4\pi)^{2}\cos(32\pi)+\frac{1}{32}(4\pi)\sin(32\pi)+\frac{1}{256}\cos(32\pi)-\left(-\frac{1}{8}(0)^{2}\cos(0)+\frac{1}{32}(0)\sin(0)+\frac{1}{256}\cos(0)\right)\ &=-\frac{1}{8}\times16\pi^{2}\times1 + 0+\frac{1}{256}\times1-(0 + 0+\frac{1}{256}\times1)\ &=- 2\pi^{2} \end{align*} ]

Answer:

$-2\pi^{2}$