Answer
See explanation
Work Step by Step
Take the substitution:
$$u=\sin(\theta)$$
The limit of integration of the integral with respect to $u$ are:
$$u=\sin(\theta) \to u=\sin(0) \to u=0$$
$$u=\sin(\theta) \to u=\sin(\frac{\pi}{6}) \to u=\frac{1}{2}$$
also
$$du=(\sin(\theta))'d\theta \to du=\cos(\theta)d\theta \to d\theta=\frac{1}{\cos(\theta)}du$$
so:
$$\int_{0}^{\frac{\pi}{6}} f(\sin(\theta))d\theta=\int_{0}^{\frac{1}{2}} f(u)\frac{1}{\cos(\theta)}du$$
Since $\cos(\theta)=\sqrt{1-\sin^{2}(\theta)}=\sqrt{1-u^{2}}$ so the above integral becomes:
$$\int_{0}^{\frac{\pi}{6}} f(\sin(\theta))d\theta=\int_{0}^{\frac{1}{2}} f(u)\frac{1}{\cos(\theta)}du=\int_{0}^{\frac{1}{2}} f(u)\frac{1}{\sqrt{1-u^{2}}}du$$
