The equation of the normal to the curve

$\because y=(1+x)^{2 y}+\cos ^{2}\left(\sin ^{-1} x\right)$

$y=e^{2 y \ln (1+x)}+\cos ^{2}\left(\cos ^{-1} \sqrt{1-x^{2}}\right)$

$=e^{2 y \ln (1+x)}+\left(1-x^{2}\right)$

$\frac{d y}{d x}=(1+x)^{2 y}\left[2 \ln (1+x) \frac{d y}{d x}+\frac{2 y}{1+x}\right]-2 x$

When $x=0$, then $y=2$

$\therefore \frac{d y}{d x}=4$

Slope of normal at $x=0$ is $-\frac{1}{4}$

$\therefore$ Equation of normal: $y-2=-\frac{1}{4}(x-0)$

$\Rightarrow x+4 y=8$