I have the following function in polar coordinates: $$f(\theta_v, \theta^´, \phi^´) = \sin \left(\theta _s\right)*\sin \left(\theta ^´\right)*\cos (\phi^´ )+\cos \left(\theta _s\right)*\cos \left(\theta ^´\right)*\cos (2 \phi^´ )$$
the function as described above should be integrated as follows: $$\int_0^{2\pi} \int_0^{\pi/2} f(\theta_v, \theta^´, \phi^´)\ * \sin \left(\theta ^´\right) d\theta ^´d\phi ^´$$
In this form the solution is clear to me. But I often read that the authors substituting $\cos \left(\theta\right) = \mu$. After this substitution the integral changes to: $$\int_0^{2\pi} \int_0^{1} f(\mu_s, \mu^´, \phi^´)\, d\mu ^´d\phi ^´$$ with $d\mu ^´d\phi ^´=\sin \left(\theta ^´\right)d\theta ^´d\phi ^´$. The changes in the integration bound is clear as well.
My question is if I substitute $\cos \left(\theta ^´\right) = \mu^´$ then the function changes to: $$f(\mu_s, \mu^´, \phi^´) = \sin \left(\theta _s\right)*\sin \left(\theta ^´\right)*\cos (\phi^´ )+\mu_s*\mu^´*\cos (2 \phi^´ )$$
How it is now possible to include the $\sin \left(\theta ^´\right)$ in the integration?
Since $\sin(\theta') \geq 0$ holds for $0 \leq \theta' \leq \frac{\pi}{2}$ , you can use $\sin^2 (x) + \cos^2 (x) = 1 \, , \, x \in \mathbb{R} \, ,$ to write $$ \sin(\theta') = \sqrt{1-\cos^2 (\theta')}= \sqrt{1-\mu'^2} \, .$$