evaluate the integral. ∫₀⁸π t² sin(2t) dt need help? read it

Answer:

$32\pi$

Explanation:

Step1: Apply integration - by - parts formula

The integration - by - parts formula is $\int u\mathrm{d}v=uv-\int v\mathrm{d}u$. Let $u = t^{2}$, $\mathrm{d}v=\sin(2t)\mathrm{d}t$. Then $\mathrm{d}u = 2t\mathrm{d}t$, $v=-\frac{1}{2}\cos(2t)$. So, $\int t^{2}\sin(2t)\mathrm{d}t=-\frac{1}{2}t^{2}\cos(2t)+\int t\cos(2t)\mathrm{d}t$.

Step2: Apply integration - by - parts again on $\int t\cos(2t)\mathrm{d}t$

Let $u = t$, $\mathrm{d}v=\cos(2t)\mathrm{d}t$. Then $\mathrm{d}u=\mathrm{d}t$, $v=\frac{1}{2}\sin(2t)$. So, $\int t\cos(2t)\mathrm{d}t=\frac{1}{2}t\sin(2t)-\frac{1}{2}\int\sin(2t)\mathrm{d}t$.

Step3: Integrate $\int\sin(2t)\mathrm{d}t$

$\int\sin(2t)\mathrm{d}t=-\frac{1}{2}\cos(2t)+C$. Combining the results, $\int t^{2}\sin(2t)\mathrm{d}t=-\frac{1}{2}t^{2}\cos(2t)+\frac{1}{2}t\sin(2t)+\frac{1}{4}\cos(2t)+C$.

Step4: Evaluate the definite integral

$\left[-\frac{1}{2}t^{2}\cos(2t)+\frac{1}{2}t\sin(2t)+\frac{1}{4}\cos(2t)\right]_0^{8\pi}$ $=-\frac{1}{2}(8\pi)^{2}\cos(16\pi)+\frac{1}{2}(8\pi)\sin(16\pi)+\frac{1}{4}\cos(16\pi)-\left(-\frac{1}{2}(0)^{2}\cos(0)+\frac{1}{2}(0)\sin(0)+\frac{1}{4}\cos(0)\right)$ Since $\cos(16\pi) = 1$ and $\sin(16\pi)=0$, we have: $=-\frac{1}{2}(64\pi^{2})(1)+\frac{1}{2}(8\pi)(0)+\frac{1}{4}(1)-\left(0 + 0+\frac{1}{4}(1)\right)$ $=- 32\pi^{2}+\frac{1}{4}-\frac{1}{4}=32\pi$.