Proving $\int_0^{\pi/2-\phi_x} \chi^*\cos\psi\sin\psi\,d\psi = \frac\pi4(1-\sin\phi_x)$, where $\cos\chi^*=\tan \phi_x \tan \psi$ I want to prove the following result
$$
I_1(\phi_x)
=\int_{0}^{\pi/2-\phi_x}
\chi^* \cos{\psi} \sin{\psi} d\psi
= \frac{\pi}{4}[1 - \sin \phi_x ]
$$
where
$\cos \chi^* = \tan \phi_x \tan \psi$.
For $\phi_x=0$,it holds
$$I_1(0)
=\frac{\pi}{2}
\int_{0}^{\pi/2}
 \cos{\psi} \sin{\psi} d\psi
=\frac{\pi}{4} \int_{0}^{\pi/2} \sin(2\psi) d\psi=
\frac{\pi}{4} [-\frac12 \cos(2\psi) ]^{\pi/2}_0 =
\frac{\pi}{4}
$$
I have done experiments in Matlab and this is true but I do not know how to proceed in the general case. Thank you in advance for your help
 A: $$\begin{align*}
I(\phi) &= \int_{0}^{\frac\pi2-\phi} \arccos\bigg(\tan(\phi) \tan(\psi)\bigg) \cos(\psi) \sin(\psi) \, d\psi \\[1ex]
&= \int_{0}^1 \arccos(\alpha) \frac{\alpha\cot^2(\phi)}{\left(1+\alpha^2\cot^2(\phi)\right)^2} \, d\alpha \tag{1} \\[1ex]
&= \int_{0}^1 \int_\alpha^1 \frac{\alpha\cot^2(\phi)}{\sqrt{1-\beta^2} \left(1+\alpha^2\cot^2(\phi)\right)^2} \, d\beta\,d\alpha\tag{2} \\[1ex]
&= \int_{0}^1 \int_0^\beta \frac{\alpha\cot^2(\phi)}{\sqrt{1-\beta^2} \left(1+\alpha^2\cot^2(\phi)\right)^2} \, d\alpha\,d\beta \tag{3} \\[1ex]
&= \int_{0}^1 \frac{\beta^2\cot^2(\phi)}{2(1+\beta^2\cot^2(\phi))\sqrt{1-\beta^2}} \,d\beta \\[1ex]
&= \frac{\cot^2(\phi)}2 \int_{0}^{\frac\pi2} \frac{\sin^2(\gamma)}{1+\sin^2(\gamma)\cot^2(\phi)} \, d\gamma \tag{4} \\[1ex]
&= \frac12 \int_{0}^{\frac\pi2} \left(1 - \frac1{1+\sin^2(\gamma)\cot^2(\phi)}\right) \, d\gamma \tag{5} \\[1ex]
&= \frac\pi4 - \frac12 \int_{0}^{\frac\pi2} \frac1{1+\sin^2(\gamma)\cot^2(\phi)} \, d\gamma \\[1ex]
&= \frac\pi4 - \frac12 \int_0^\infty \frac1{1+\delta^2\csc^2(\phi)} \, d\delta \tag{6} \\[1ex]
&= \frac\pi4 - \frac12 \sin(\phi) \int_0^{\frac\pi2} d\epsilon \tag{7} \\[1ex]
&= \boxed{\frac\pi4 (1-\sin(\phi))}
\end{align*}$$


*

*$(1)$ : substitute $\alpha=\tan(\phi) \tan(\psi)$

*$(2)$ : integral definition of $\arccos$

*$(3)$ : reverse order of integration

*$(4)$ : substitute $\beta=\sin(\gamma)$

*$(5)$ : partial fractions (optional)

*$(6)$ : substitute $\delta=\tan(\gamma)$

*$(7)$ : substitute $\delta=\sin(\phi)\tan(\epsilon)$
