Answer
$$\int\frac{\sin\sqrt\theta}{\sqrt{\theta\cos^3\sqrt\theta}}d\theta=\frac{4}{\sqrt{\cos\sqrt\theta}}+C$$
Work Step by Step
$$A=\int\frac{\sin\sqrt\theta}{\sqrt{\theta\cos^3\sqrt\theta}}d\theta$$
Set $u=\cos\sqrt\theta$, then $$du=-\sin\sqrt\theta\times(\sqrt\theta)'d\theta=-\frac{\sin\sqrt\theta}{2\sqrt\theta}d\theta$$
That means, $$\frac{\sin\sqrt\theta}{\sqrt\theta}d\theta=-2du$$
Therefore,
$$A=-2\int\frac{1}{\sqrt{u^3}}du=-2\int u^{-3/2}du$$ $$A=-2\times\frac{u^{-1/2}}{-\frac{1}{2}}+C=4u^{-1/2}+C=\frac{4}{\sqrt u}+C$$ $$A=\frac{4}{\sqrt{\cos\sqrt\theta}}+C$$
