Big Integrals

Note that $\int_0^{\pi/2} \cos^2(\cos x)\, dx = \int_0^{\pi/2} \cos^2\left(\sin \left(\frac{\pi}{2} - x\right)\right)\, dx = \int_{\pi/2}^0 \cos^2(\sin u)\, (-du) = \int_0^{\pi/2} \cos^2(\sin u)\,du$

Therefore, if we write that last integral using $x,$ we can use write $\int_0^{\pi/2} \Big(\cos^2(\cos x) + \sin^2(\sin x)\Big)\, dx = \int_0^{\pi/2} \Big(\cos^2(\sin x) + \sin^2(\sin x)\Big)\, dx = \int_0^{\pi/2} \Big(1\Big)\, dx = \frac{\pi}{2}$ giving the answer of $A = \boxed{2}$

$\begin{aligned} I & = \int_0^\frac \pi 2 \left(\cos^2(\cos x) + \sin^2 (\sin x)\right) dx & \small \color{#3D99F6} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\cos^2(\cos x) + \sin^2 (\sin x) + \cos^2\left(\cos \left(\frac \pi 2 - x\right) \right) + \sin^2 \left(\sin \left(\frac \pi 2 - x\right) \right) \right) dx & \small \color{#3D99F6} \text{Since }\sin \left(\frac \pi 2- \theta \right) = \cos \theta \text{ and }\cos \left(\frac \pi 2 - \theta \right) = \sin \theta \\ & = \frac 12 \int_0^\frac \pi 2 \left({\color{#3D99F6}\cos^2(\cos x)} + \color{#D61F06}\sin^2 (\sin x) + \cos^2 (\sin x) + \color{#3D99F6} \sin^2 (\cos x) \right) dx & \small \color{#3D99F6} \text{and } \sin^2 \theta + \cos^2 \theta = 1 \\ & = \frac 12 \int_0^\infty 2 \ dx = \int_0^\infty \ dx = \frac \pi 2 \end{aligned}$

Therefore, $A = \boxed 2$ .