Integrate challenge 1

This question revolves around a couple of identities.

_Identity 1: _ $\prod_{r=0}^{n-1} \sin \left( x + \frac{r \pi }{n} \right) = \frac{\sin (nx)}{2^{n-1}}$

Proof:

Consider the equation, $z^n = e^{2nix}$ . We can factorize the above equation using the concept of roots of unity.
$\displaystyle \begin{array}{c}\\ z^n - e^{2nix} && = \prod_{r=0}^{n-1} \left( z - e^{i \left(2x + \frac{2 \pi r}{n}\right) } \right) \\ | 1 - e^{2nix} | && = \prod_{r=0}^{n-1} \bigg| \left( 1 - e^{i \left(2x + \frac{2 \pi r}{n}\right) } \right) \bigg| \\ 2 \sin (nx) \big| \sin (nx) -i \cos (nx) \big| && = \prod_{r=0}^{n-1} 2 \sin \left(x + \frac{\pi}{n} \right) \bigg| \sin \left(x + \frac{\pi}{n} \right) -i \cos \left(x + \frac{\pi}{n}\right) \bigg| \\ 2 \sin (nx) && = 2^{n} \prod_{r=0}^{n-1} \sin \left( x + \frac{r \pi }{n} \right) \\ \end{array}$

And the identity follows.

Identity 2: $\cos ^2 A - \cos ^2 B = \sin (B+A) \sin (B - A)$

Proof:

$\displaystyle \begin{array}{c}\\ \sin (B+A) \sin (B - A) && = ( \sin A \cos B + \cos A \sin B)( \cos A \sin B - \sin A \cos B) \\ && = \cos^2 A \sin^2 B - \sin^2 A \cos^2 B \\ && = \cos^2 A - \cos^2 A \cos^2 B - \sin^2 A \cos^2 B \\ && = \cos ^2 A - \cos ^2 B \\ \end{array}$

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Using the second identity, we can write, $\displaystyle \sin x ( \cos ^2 x - \cos ^2 y )\sin x ( \cos ^2 x - \cos ^2 2y )$

$\displaystyle \begin{array}{c}\\ && = \sin x \sin (x + y ) \sin (y - x) \sin (x + 2y) \sin (2y - x) \\ && = \sin \left( x - \frac{2 \pi }{5} \right) \sin \left( x - \frac{ \pi }{5} \right) \sin (x) \sin \left( x + \frac{ \pi }{5} \right) \sin \left( x + \frac{2 \pi }{5} \right) \\ && = \sin (x) \sin \left( x + \frac{ \pi }{5} \right) \sin \left( x + \frac{2 \pi }{5} \right) \sin \left( x + \frac{3 \pi }{5} \right) \sin \left( x + \frac{ 4\pi }{5} \right) \\ && = \frac{ \sin(5x) }{2^4} \\ \end{array}$

The last step follows from the first identity. Therefore,

$\displaystyle \int_0^1 \frac{ \sin x ( \cos ^2 x - \cos ^2 y )\sin x ( \cos ^2 x - \cos ^2 2y )}{ \sin (5x) } \text{ d}x$
$\displaystyle = \int_0^1 \frac{ \sin (5x)}{16}. \frac{1}{ \sin (5x) } \text{ d}x = \int_0^1 \frac{1}{16} \text{ d}x = \frac{1}{16}$

Compare the answer to find the values of $a$ and $b$ and add them up to get the final answer as $\displaystyle \boxed{17}$ .