Given a manifold ${\cal M}$, we define a function on ${\cal M}$: $$ f(x)=h(d(x,\alpha)),\quad x\in{\cal M} $$ where $\alpha\in{\cal M}$ is a point, $d(\cdot,\cdot)$ is the geodesic distance, and $h:\mathbb{R}\to\mathbb{R}$ is a function. Consider the integral $$ \int_{\cal M}f(x)dx. $$ When ${\cal M}=\mathbb{S}^{m}$ is a sphere, this integral can be rewritten as $$ {\rm vol}(\mathbb{S}^{m-1})\int_0^\pi h(r)\sin^{m-1}(r)dr. $$ When ${\cal M}=\mathbb{C}\mathbb{P}^m$ is a complex projective space, can I rewrite this integral in a similar way?
The following is my own answer. Maybe someone could provide some intuitive explanations.
When ${\cal M}=\mathbb{C}\mathbb{P}^m$ is a complex projective space, we may use the following coordinate representation. For any $[x]\in \mathbb{C}\mathbb{P}^{m}$, we can choose a representation element as $$ x=(z_0,z_1,\ldots,z_m)=(a_0,a_1e^{i\nu_1},\ldots,a_ne^{i\nu_m})\in\mathbb{C}^{m+1}, $$ where $a_i\geq 0$, $\sum a_i^2=1$, and $\nu_i\in[0,2\pi)$. Furthermore, we define $$\left\{ \begin{align} a_0&=\sin \theta_1\sin\theta_2\cdots\sin\theta_m,\\ a_1&=\cos\theta_1\sin\theta_2\cdots \sin\theta_m,\\ a_2&=\cos\theta_2\sin\theta_3\cdots\sin\theta_m,\\ &\ldots\\ a_m&=\cos \theta_m, \end{align}\right. $$ where $\theta_i\in(0,\pi/2)$. This gives us a coordinate representation of $\mathbb{C}\mathbb{P}^m$ using $\{(\theta_1,\ldots,\theta_m,\nu_1,\ldots,\nu_m)\mid \theta_i\in(0,\pi/2),\nu_i\in(0,2\pi)\}$. Using this coordinate representation, the volume form is $$ {\rm dvol}=\prod_{i=1}^m\cos\theta_i\sin^{2i-1}\theta_id\theta_id\nu_i. $$ See section 4.6 and 4.7 of [1] for more details.
Now we proceed to evaluate the integral $I=\int_{\cal M}h(d(x,\alpha))dx$. Since $\cal M$ is homogeneous, we assume without loss of generality that $\alpha=(0,\ldots,0,1)\in\mathbb{C}^{m+1}$. Notice that the geodesic distance satisfies $$ d(x,\alpha)=\arccos(|\langle x,\alpha\rangle|)=\arccos (\cos\theta_m)=\theta_m. $$ Then we have $$ \begin{align} I&=\int \prod_{i=1}^{m-1}\cos\theta_i\sin^{2i-1}\theta_id\theta_id\nu_i \cdot \int_0^{2\pi} d\nu_m \int_0^{\pi/2} h(\theta_m)\cos\theta_m\sin^{2m-1}\theta_md\theta_m\\ &={\rm vol}(\mathbb{C}\mathbb{P}^{m-1})\cdot 2\pi\int_0^{\pi/2}h(r)\cos r\sin^{2m-1}rdr. \end{align} $$
[1] I. Bengtsson, K. Życzkowski (2017). Geometry of quantum states: an introduction to quantum entanglement.