Answer
$$-\sin{\frac{1}{\theta}}+C$$
Work Step by Step
$\int\frac{1}{\theta^2}\cos{\frac{1}{\theta}}d\theta$
Let $u=\frac{1}{\theta}\hspace{8mm}du=-\frac{1}{\theta^2}d\theta\hspace{8mm}d\theta=-\theta^2du$
$\int\frac{1}{\theta^2}\cos{\frac{1}{\theta}}d\theta=\int\frac{1}{\theta^2}\cos{u}(-\theta^2)du=\int-\cos{u}du=-\sin{u}+C$
$=-\sin{\frac{1}{\theta}}+C$
