Answer
$$\int\frac{1}{\theta^2}\sin\frac{1}{\theta}\cos\frac{1}{\theta}d\theta=\frac{1}{4}\cos\frac{2}{\theta}+C$$
Work Step by Step
$$A=\int\frac{1}{\theta^2}\sin\frac{1}{\theta}\cos\frac{1}{\theta}d\theta$$
We set $u=\frac{1}{\theta}$
Then $$du=-\frac{1}{\theta^2}d\theta$$
That means $$\frac{1}{\theta^2}d\theta=-du$$
Therefore, $$A=-\int\sin u\cos udu=-\frac{1}{2}\int2\sin u\cos udu$$
Apply the identity $2\sin u\cos u=\sin2u$
$$A=-\frac{1}{2}\int\sin2udu$$ $$A=-\frac{1}{2}\times\Big(-\frac{1}{2}\cos2u\Big)+C=\frac{1}{4}\cos2u+C$$ $$A=\frac{1}{4}\cos\frac{2}{\theta}+C$$
