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Solution:

This problem can be solved using Trig Sub, and Beta's Integral.

First step is let $x = \sin \theta \rightarrow \sqrt{1-x^2} = \cos \theta$ .

This will cause the bounds to be changed: From $x: [0, 1]$ to $\theta: \left[0, \dfrac{\pi}{2}\right]$ .

From the first step, $dx = \cos \theta d\theta \Longrightarrow d\theta = \dfrac{dx}{\sqrt{1-x^2}}$ .

Manipulating the integrand into: $\displaystyle \int_0^1 \left(3x - 4x^3\right)^{2015} * \left[\sqrt{1-x^2}*\left(1-4x^2\right)\right]^{1991} \dfrac{dx}{\sqrt{1 - x^2}}$ and substituting the Trig Sub we initiate earlier:

$\int_0^\frac{π}{2} \left(3\sin\theta - 4\sin^3 \theta\right)^{2015} * \left[\sqrt{1-\sin^2 \theta} * \left(1 - 4\sin^2 \theta\right)\right]^{1991} d\theta$

and knowing that $3\sin\theta - 4\sin^3 \theta = \sin 3\theta$ ; $\sqrt{1-\sin^2 \theta} = \cos\theta$ ; and manipulating $1-4\sin^2 \theta = \cos^2 \theta - 3\sin^2 \theta$ :

$\int_0^\frac{π}{2} \left(\sin 3\theta\right)^{2015} * \left[\cos\theta * \left(\cos^2 \theta - 3\sin^2 \theta\right)\right]^{1991} d\theta$

Using the angle identity derived from De Moivre's theorem $\cos^3 \theta - 3\cos\theta\sin^2 \theta = \cos 3\theta$ , the integral will become:

$\int_0^\frac{π}{2} \left(\sin 3\theta\right)^{2015} * \left(\cos 3\theta\right)^{1991} d\theta$

Now, letting $u = \sin^2 3\theta \rightarrow 1-u = \cos^2 3\theta$ .

This will cause again the bounds to shift from $\theta: \left[0, \dfrac{π}{2}\right]$ to $u: \left[0, 1\right]$ .

From there, $du = 2\sin 3\theta \cos 3\theta * 3 d\theta = 6u^{0.5} \left(1-u\right)^{0.5} d\theta$ . Thus,

$d\theta = \dfrac{1}{6} * u^{-0.5} * \left(1 - u\right)^{-0.5} du$

The integrand will become then:

$\dfrac{1}{6} \int_0^1 u^{\frac{2015}{2}} * \left(1 - u\right )^{\frac{1991}{2}} * u^{-0.5} * \left(1-u\right)^{-0.5} du \\ = \dfrac{1}{6} \int_0^1 u^{1007} \left(1-u\right)^{995} du$

And this is an example of a beta Integral. It can be evaluated as:

$\dfrac{1}{6} \left(\dfrac{\Gamma(1007+1) * \Gamma(995+1) }{\Gamma(1007+995+2)} \right) \\ = \dfrac{1}{6} \left(\dfrac{1007! * 995!}{2002!} \right) * \dfrac{1}{2003} \\ = \dfrac{1}{12018*C(2002, 995)}$

Hence, the expression in the problem above is equivalent to: $\displaystyle \boxed{12018*C(2002, 995)}$

And therefore, $a + b = 12018 + 2002 = \boxed{14020}$ .