A calculus problem by Rudraksh Shukla

$\begin{aligned} \int_0^\frac \pi 2 (\sin x + a\cos x)^3 dx - \frac {4a}{\pi -2} \color{#3D99F6} \int_0^\frac \pi 2 x\cos x\ dx & = 2 & \small \color{#3D99F6} \text{By integration by parts} \\ \int_0^\frac \pi 2 \left(1+a^2\right)^\frac 32 \sin^3 ({\color{#D61F06}x + \phi}) dx - \frac {4a}{\pi -2} \color{#3D99F6} \left(x \sin x \bigg|_0^\frac \pi 2 - \int_0^\frac \pi 2 \sin x\ dx \right) & = 2 & \small \color{#D61F06} \text{where }\phi = \tan^{-1} a \\ \left(1+a^2\right)^\frac 32 \int_0^\frac \pi 2 \sin^3 \left({\color{#D61F06}\frac \pi 2 - x + \phi}\right) dx - \frac {4a}{\pi -2} \left(\frac \pi 2 + \cos x \ \bigg|_0^\frac \pi 2 \right) & = 2 & \small \color{#D61F06} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ \left(1+a^2\right)^\frac 32 \int_0^\frac \pi 2 \cos^3 (x - \phi) dx - \frac {4a}{\pi -2} \left(\frac \pi 2 -1 \right) & = 2 & \small \color{#3D99F6} \text{Let } \theta = x - \phi \implies d\theta = dx \\ \left(1+a^2\right)^\frac 32 \int_{-\phi}^{\frac \pi 2-\phi} \cos^3 \theta \ d\theta - 2a & = 2 \\ \left(1+a^2\right)^\frac 32 \int_{-\phi}^{\frac \pi 2-\phi} (1-\sin^2 \theta) \cos \theta \ d\theta - 2a & = 2 & \small \color{#3D99F6} \text{Let } u = \sin \theta \implies du = \cos \theta \ d\theta \\ \left(1+a^2\right)^\frac 32 \int_{-\sin \phi}^{\cos \phi} (1-u^2) \ du - 2a & = 2 \\ \left(1+a^2\right)^\frac 32 \left[u - \frac {u^3}3 \right]_{-a/\sqrt{1+a^2}}^{1/\sqrt{1+a^2}} - 2a & = 2 \\ \left(1+a^2\right)^\frac 32 \left[\frac {1+a}{\sqrt{1+a^2}} - \frac {1+a^3}{3\left(1+a^2\right)^\frac 32} \right] - 2a & = 2 \\ (1+a^2)(1+a) - \frac 13(1+a^3) - 2a - 2 & = 0 \\ 2x^3+3x^2-3x-4 & = 0 \\ (x+1)(2x^2+x-4) & = 0 \end{aligned}$

Let $a_1 = - 1$ , then by Vieta's formula $a_2+a_3 = - \dfrac 12$ and $a_2a_3 = -2$ . Then we have:

$\begin{aligned} a_1^2 + a_2^2 + a_3^2 & = 1^2 + (a_2+a_3)^2 - 2a_2a_3 \\ & = 1 + \frac 14 + 4 = 5.25 \\ \implies 1000(a_1^2 + a_2^2 + a_3^2) & = \boxed{5250} \end{aligned}$