Answer
$$\displaystyle\int t\sin{t}\cos{t}\thinspace dt=\frac{\sin{2t}}{8}-\frac{t\cos{2t}}{4}+C$$
Work Step by Step
Use the half angle trignometric identity $\sin{2\theta}=2\sin{\theta}\cos{\theta}$ to rewrite $\sin{\theta}\cos{\theta}=\frac{\sin{2\theta}}{2}$
$\displaystyle\int t\sin{t}\cos{t}\thinspace dt = \displaystyle\int t\frac{\sin{2t}}{2}\thinspace dt$
Pull the constant outside $\frac{1}{2}\int t\sin{2t}\thinspace dt$
Integration by parts: $uv-\int v\thinspace du$, let $u =t$ and $dv=\sin(2t)dt$ then, $du=dt$ and $v=\big(\frac{1}{2}\big)(-\cos{2t})$ $$=\frac{1}{2}\bigg[t\bigg(\frac{1}{2}\bigg)(-\cos{2t})-\int\bigg(\frac{1}{2}\bigg)(-\cos{2t})\thinspace dt\bigg]$$ $$=\frac{1}{2}\bigg[t\bigg(\frac{1}{2}\bigg)(-\cos{2t})+\int\bigg(\frac{1}{2}\bigg)(\cos{2t})\thinspace dt\bigg]$$ $$=\frac{1}{2}\Bigg[t\bigg(\frac{1}{2}(-cos{2t})+\bigg(\frac{1}{2}\bigg)\bigg(\frac{1}{2}\bigg)\sin{2t}+C\Bigg]$$ $$=\frac{1}{2}\Bigg[\bigg(\frac{-t\cos{2t}}{2}\bigg)+\bigg(\frac{\sin{2t}}{4}\bigg)+C\Bigg]$$ $$=\bigg(\frac{-tcos{2t}}{4}\bigg)+\bigg(\frac{\sin{2t}}{8}\bigg)+C$$ $$=\frac{\sin{2t}}{8}-\frac{t\cos{2t}}{4}+C$$
