I know that $$\frac{d}{dx} \sinh^{-1}x=\frac{1}{\sqrt{1+x^2}}$$ so I wanted to use that along with the chain rule to integrate $$\int\frac{1}{\sqrt{1+\cos^2x}}dx$$$$ = \sinh^{-1}x(\cos^2x)(2\cos x)(-\sin x) + C = -2\sinh^{-1}x\cos^3x\sin x + C$$ but WolframAlpha says you instead need to use the elliptic integral of the first kind.
Why is that necessary?
You can't use the chain rule in that way when integrating, only differentiating. In other words, $$\int f(g(x)) dx \ne \int f(x) dx \times g'(x).$$
The chain rule (in the way you're using it) only works with differentiation, that is, $\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x)$. You can only say that $$ \frac{d}{dx} \sinh^{-1} ( \cos{x}) = \frac{-\sin{x}}{\sqrt{1+\cos^2{x}}} \Rightarrow \int \frac{-\sin{x}}{\sqrt{1+\cos^2{x}}} = \sinh^{-1} ( \cos{x}) + C.$$
Note that in general there is a "reverse chain rule" that you can use: $$ \int g'(x) f(g(x)) dx = F(g(x)) + C,$$ where $F$ is the antiderivative of $f$. So in the example of $\int \frac{1}{\sqrt{1+(2x)^2}} dx = \frac{1}{2} \sinh^{-1} (2x) + C$, what is going on is that $g(x)$ is a linear function so $g'(x)$ is a constant which you can move out of the integral. However, in your original example, $g(x)$ would have to be $\cos x$, so $g'(x) = - \sin x$, so the reverse chain rule isn't helpful here.
To solve this particular integral, it turns out it is in the form of an elliptical integral of the first kind. If you use the fact that $\sin^2 x + \cos^2 x = 1$, your integral becomes: $$ \int \frac{1}{\sqrt{2-\sin^2 x}} dx = \frac{1}{\sqrt{2}} F\left(x \; \middle| \; \frac{1}{2} \right),$$ where $F(x | k^2) = \int_0^x \frac{d \theta}{\sqrt{1 - k^2 \sin^2 \theta}}$