Answer
$$\int\frac{\cos\sqrt\theta}{\sqrt\theta\sin^2\sqrt\theta}d\theta=-\frac{2}{\sin\sqrt{\theta}}+C$$
Work Step by Step
$$A=\int\frac{\cos\sqrt\theta}{\sqrt\theta\sin^2\sqrt\theta}d\theta$$
We set $u=\sin\sqrt\theta$
Then $$du=(\sqrt\theta)'\cos\sqrt\theta d\theta=\frac{\cos\sqrt\theta}{2\sqrt\theta}d\theta$$
That means $$\frac{\cos\sqrt\theta}{\sqrt\theta}d\theta=2du$$
Therefore, $$A=2\int\frac{1}{u^2}du=-2\int-\frac{1}{u^2}du$$ $$A=-2\times\frac{1}{u}+C=-\frac{2}{u}+C$$ $$A=-\frac{2}{\sin\sqrt{\theta}}+C$$
