Let $$f: \mathbb{R} \to \mathbb{R}$$ be a function defined by $$f(x) = \frac{4^x}{4^x + 2}$$ and $$M = \int_{f(a)}^{f(1-a)} x\sin^4(x(1-x))dx$$, $$N = \int_{f(a)}^{f(1-a)} \sin^4(x(1-x))dx; a \neq \frac{1}{2}$$. If $$\alpha M = \beta N, \alpha, \beta \in \mathbb{N}$$, then the least value of $$\alpha^2 + \beta^2$$ is equal to ______

We are given $$f(x)=\dfrac{4^{x}}{4^{x}+2}$$ and the two integrals
$$M=\int_{f(a)}^{f(1-a)} x\,\sin^{4}\!\bigl(x(1-x)\bigr)\,dx, \qquad N=\int_{f(a)}^{f(1-a)} \sin^{4}\!\bigl(x(1-x)\bigr)\,dx,$$ with $$a\neq\dfrac12.$$ We must find the least natural numbers $$\alpha,\beta$$ satisfying $$\alpha M=\beta N$$ and then evaluate $$\alpha^{2}+\beta^{2}.$$

Step 1: Show that the limits are symmetric about $$\dfrac12$$.
Write $$A=4^{a}.$$ Then $$f(a)=\dfrac{A}{A+2},\qquad f(1-a)=\dfrac{4^{1-a}}{4^{1-a}+2}=\dfrac{4/A}{4/A+2}=\dfrac{4}{4+2A} =\dfrac{2}{A+2}.$$ Hence $$f(a)+f(1-a)=\dfrac{A}{A+2}+\dfrac{2}{A+2}=1.$$ Therefore $$\boxed{\,f(1-a)=1-f(a)\,}.$$ Put $$p=f(a)\;(0\lt p\lt1,\;p\neq\tfrac12).$$ Then the limits of both integrals are $$p\;\text{ to }\;1-p,$$ which are equidistant from $$\dfrac12.$$

Step 2: Use symmetry of the integrand.
Define $$g(x)=x(1-x).$$ Clearly $$g(1-x)=g(x),$$ so $$F(x)=\sin^{4}\!\bigl(g(x)\bigr)$$ satisfies $$F(1-x)=F(x),$$ i.e. $$F(x)$$ is symmetric about $$x=\dfrac12.$$

Step 3: Express the integrals around the centre $$\dfrac12$$.
Let $$x=\dfrac12+y \quad\Longrightarrow\quad y=x-\dfrac12.$$ Because the interval $$[\,p,\,1-p\,]$$ is symmetric about $$\dfrac12,$$ we have $$p=\dfrac12-d,\;\;1-p=\dfrac12+d$$ for some $$d\;(0\lt d\lt\dfrac12).$$ Then $$$ \begin{aligned} N &=\int_{-d}^{\,d} F\!\Bigl(\tfrac12+y\Bigr)\,dy,\\ M &=\int_{-d}^{\,d} \Bigl(\tfrac12+y\Bigr)\,F\!\Bigl(\tfrac12+y\Bigr)\,dy. \end{aligned} $$$

Step 4: Evaluate $$M$$ in terms of $$N$$.
Because $$F\!\bigl(\tfrac12+y\bigr)$$ is an even function of $$y,$$ denote it by $$E(y).$$ Then $$N=\int_{-d}^{\,d} E(y)\,dy.$$ Now, $$$ \begin{aligned} M &=\int_{-d}^{\,d} \Bigl(\tfrac12+y\Bigr)E(y)\,dy \\ &=\tfrac12\!\int_{-d}^{\,d} E(y)\,dy \;+\;\int_{-d}^{\,d} y\,E(y)\,dy. \end{aligned} $$$ The second integral vanishes because $$yE(y)$$ is an odd function integrated over $$[-d,d].$$ Hence $$\boxed{\,M=\dfrac12\,N\,}.$$

Step 5: Relate $$\alpha$$ and $$\beta$$.
Given $$\alpha M=\beta N,$$ substitute $$M=\dfrac12N$$: $$$ \alpha\Bigl(\dfrac12 N\Bigr)=\beta N \;\;\Longrightarrow\;\; \dfrac{\alpha}{2}=\beta \;\;\Longrightarrow\;\; \boxed{\alpha=2\beta}. $$$

Step 6: Minimise $$\alpha^{2}+\beta^{2}$$ for natural $$\alpha,\beta.$$
Take the smallest natural $$\beta=1,$$ then $$\alpha=2.$$ Thus $$\alpha^{2}+\beta^{2}=2^{2}+1^{2}=4+1=5.$$

The least possible value of $$\alpha^{2}+\beta^{2}$$ is therefore 5.