<s>[INST] If $a_{1}\;a_{2},a_{3}$ are the Roots of cubic eq. , Then $1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right)$ 
If $a_{1}\;a_{2},a_{3}$ are three real values of $a$ which satisfy the equation $$\displaystyle \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx-\frac{4a}{\pi-2}\int_{0}^{1}x\cdot \cos xdx = 2.$$ Then value of $\displaystyle 1000\left(a^2_{1}+a^2_{2}+a^2_{3}\right) = $

$\bf{My\; Trial::}$ Let $\displaystyle I = \int_{0}^{1}\left(\sin x+a\cdot \cos x\right)^3dx=\int_{0}^{1}\left(\sin^3 x+a^3\cos^3 x+3a\sin^2 x\cdot \cos x+3a^2\sin x\cdot \cos^2 x \right)dx$
So $\displaystyle I = -\int_{0}^{1}(1-\cos^2 x)\cdot (\cos x)^{'}dx+a^3\int_{0}^{1}(1-\sin^2 x)\cdot (\sin x)^{'}dx+3a\int_{0}^{1}(\sin x)^2\cdot (\sin x)^{'}dx-3a^2\int_{0}^{1} (\cos x)^2\cdot (\cos x)^{'}dx$
and Let $\displaystyle J = \int_{0}^{1}x\cdot \cos xdx = \left[x\cdot \sin x+\cos x\right]_{0}^{1} = \left(\sin 1+\cos 1-1\right)$
Now How can I solve after that
Help me
Thanbks
 [/INST] Continuing from JimmyK4542's answer, we find that $$C_3 = \displaystyle\int_0^1 \cos^3 x\,dx=\frac{1}{12} (9 \sin (1)+\sin (3))$$ $$C_2 = \displaystyle\int_0^1 3\sin x \cos^2 x\,dx=1-\cos ^3(1)$$ $$C_1 = \displaystyle\int_0^1 3\sin^2 x \cos x\,dx - \dfrac{4}{\pi - 2}\int_0^1 x\cos x\,dx=\sin ^3(1)-\frac{4 (-1+\sin (1)+\cos (1))}{\pi -2}$$ $$C_0 =\displaystyle-2+\int_0^1 \sin^3 x\,dx=\frac{1}{12} (-16-9 \cos (1)+\cos (3))$$ and $$a_1^2+a_2^2+a_3^2 = (a_1+a_2+a_3)^2-2(a_1a_2+a_2a_3+a_3a_1) = \dfrac{C_2^2-2C_1C_3}{C_3^2}$$ can be computed. However, the result is far to be simple; the numerical value is $4.02458872035$.
But, as I said in my comment, two of the roots of the cubic equation are complex $$a_1=-1.29971 + 0.712738 i$$ $$a_2=1.29971 - 0.712738 i$$ $$a_3=1.28923$$
Could it be possible that there are typo's in the expression ?
</s>