Consider the series  $f\left(x\right)=1+3{\sin }^{2}x+5{\sin }^{4}x+\mathrm{.....}\infty ,g\left(x\right)=\cos x+\cos 3x+\cos 5x+$ …..n terms, then

 $f\left(x\right)={\sec }^{2}x+2{\sec }^{2}x{\tan }^{2}x$
$g\left(x\right)=\frac{\sin \left(2nx\right)}{2\sin x}$
 A)  $\int \left(1+2{\tan }^{2}x\right){\sec }^{2}xdx$$=\tan x+\frac{2}{3}{\tan }^{3}x+C$ 
B)  $\underset{n\to 0}{\lim }\frac{\left({\sec }^{2}x+2{\sec }^{2}x{\tan }^{2}x-1\right).4}{{\sin }^2.2nx}=e^{\frac{3}{n^2}}$
C)  $I_{2n+2}-I_{2n}=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin \left(2n+2\right)x-\sin 2nx}{\sin x}$

$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \left(2n+1\right)xdx=\frac{1}{\left(2n+1\right)}\sin \left(2n+1\right)\frac{\pi }{2}$

$=\frac{1}{3}.\frac{1}{5}.-\frac{1}{7}.\frac{1}{9}.\frac{-1}{11}.\frac{1}{13}forn\in \left[1,6\right]$

 D)  $g\left(x\right)=0\Rightarrow x=\frac{\pi }{2}$

$\sin 8x=0,\sin x\ne 0$

$\Rightarrow x=\frac{\pi }{8}.\frac{\pi }{4}.\frac{3\pi }{8}.\frac{\pi }{2}.\frac{5\pi }{8}.\frac{3\pi }{4}.\frac{7\pi }{8}$