Best fit line to a function

$\large \displaystyle I(m) = \int_0^{\frac{\pi}{2}} \left (mx - \frac{\sin(x)}{x} \right )^2 dx$

$\large \displaystyle I(m) = \int_0^{\frac{\pi}{2}} \left (m^2 x^2 - 2m\sin(x) + \frac{\sin^2(x)}{x^2} \right ) dx$

We're looking for the $m$ that minimizes this integral. So:

$\large \displaystyle \frac {dI(m)}{dm} = 0$

$\large \displaystyle \int_0^{\frac{\pi}{2}} \left ( 2mx^2 - 2\sin(x) \right ) dx = 0$

$\large \displaystyle \left [\frac{2mx^3}{3} + 2\cos(x) \right ]_0^{\frac{\pi}{2}} = 0$

$\large \displaystyle \frac{m\pi^3}{12} - 2 = 0$

$\color{#20A900} \boxed{ \large \displaystyle m = \frac{24}{\pi^3} }$

Thus:

$\color{#3D99F6} \large \displaystyle a = 24, b = 1, c = 3, \boxed{ \large \displaystyle a+b+c = 28}$

We are presumably trying to minimize the following integral:

$\large{I = \int_0^\frac{\pi}{2} (mx - \frac{sinx}{x})^2 dx = \int_0^\frac{\pi}{2} (m^2 x^2 - 2msinx + \frac{sin^2x}{x^2}) dx }$

We're going to differentiate this integral with respect to $m$ and set to zero, so any term in the integrand without an $m$ term is irrelevant. The integral therefore becomes:

$\large{I' = \int_0^\frac{\pi}{2} (m^2 x^2 - 2msinx) dx = \frac{\pi^3}{24} m^2 - 2m }$

Differentiating and setting to zero gives:

$\large{\frac{\pi^3}{12} m - 2 = 0 \\ m = \boxed{\frac{24}{\pi^3}}}$

this can be solved by partial derivatives but i used projection matrices. For an unsolvable system of equations the best solution is given by $A^T Ax=A^T b$ where $A$ was the original input, in this case a column vector of all numbers in $\left(0,\dfrac{\pi}{2}\right)$ . x is a scalar in this case as there is only one variable to solve in. and b is the output, in this case $\dfrac{\sin(a)}{a}$ . written in dot product notation we have $x=\dfrac{A.b}{A.A} =\dfrac{\int_0^{\pi/2} a \dfrac{\sin(a)}{a} \ da}{\int_0^{\pi/2} a a\ da}$ both are easy integrals to solve and we have $x=\dfrac{24}{\pi^3}$