Answer
$$\int \theta\cos\pi\theta d\theta=\frac{\theta\sin\pi\theta}{\pi}+\frac{\cos\pi\theta}{\pi^2}+C$$
Work Step by Step
$$A=\int \theta\cos\pi\theta d\theta$$
Take $u=\theta$ and $dv=\cos\pi\theta d\theta$
We then have $du=d\theta$ and $v=\frac{1}{\pi}\sin\pi\theta$
Apply the formula $\int udv= uv-\int vdu$, we have $$A=\theta\Big(\frac{1}{\pi}\sin\pi\theta\Big)-\int\frac{1}{\pi}\sin\pi\theta d\theta$$
$$A=\frac{\theta\sin\pi\theta}{\pi}-\frac{1}{\pi}\int\sin\pi\theta d\theta$$
$$A=\frac{\theta\sin\pi\theta}{\pi}-\frac{1}{\pi}\Big(-\frac{1}{\pi}\cos\pi\theta\Big)+C$$
$$A=\frac{\theta\sin\pi\theta}{\pi}+\frac{\cos\pi\theta}{\pi^2}+C$$
