Assertion (A) $\int\limits_{-a}^a f(x) d x=\int_0^a(f(x)+f(-x)) d x$

Reason (R) $\int\limits_a^b f(x) d x=\int_{g(a)}^{g(b)} f(g(u)) g^{\prime}(u) d u$

The correct option among the following is

If $\cos x+\cos 2 x+\ldots+\cos n x=\frac{A(x)}{2 \sin x / 2}$, then $\int\limits_0^\pi A(x) d x=$

$$ \begin{array}{r}\mathop {\lim }\limits_{n \to \infty }\left[\frac{n^{3 / 2}}{n^{5 / 2}}-\frac{n^{1 / 2}}{n^{3 / 2}}+\frac{n^{3 / 2}}{(n+2)^{5 / 2}}-\frac{n^{1 / 2}}{(n+3)^{3 / 2}}\right. \\ +\frac{n^{3 / 2}}{(n+4)^{5 / 2}}-\frac{n^{1 / 2}}{(n+6)^{3 / 2}}+\ldots+\frac{n^{3 / 2}}{(n+2(n-1))^{5 / 2}} \\ \left.-\frac{n^{1 / 2}}{(n+3(n-1))^{3 / 2}}\right]= \end{array} $$

$$ \lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]= $$