$$\int^4_{-4}\cos{x}\delta(\sin{x}) \,{\rm d} x$$
I used the identity
$$\delta(\sin{x}) = \sum\frac{\delta(x-n)}{\frac{d}{dx_i}\sin{x}}$$
Then
$$\int^4_{-4}\cos{x}\,{\rm d} x (\delta(x)+\delta(x-\pi)+\delta(n+\pi))$$
And now I am stuck. Can this be directly computed?
On
Note that formally (and in distribution sense), we have
$$ H'(x) = \delta(x) \qquad\text{where}\quad H(x) = \mathbf{1}_{[0,\infty)}(x) = \begin{cases} 1, & x \geq 0 \\ 0, & x < 0 \end{cases} $$
So, we expect that $\delta(f(x))$ is related to $H(f(x))$ via the chain rule:
$$ [H(f(x))]' = H'(f(x))f'(x) = \delta(f(x))f'(x). $$
Consequently, with $f(x) = \sin x$,
$$ \int_{-4}^{4} \cos x \, \delta(\sin x) \, \mathrm{d}x = \int_{-4}^{4} [H(\sin x)]' \, \mathrm{d}x = \left[ H(\sin x) \right]_{-4}^{4}. $$
You can then evaluate this by noting that $\sin 4 < 0$ to conclude that the integral is equal to $-1$.
The composition formula should be $$ \delta(g(x))=\sum_{x_i:g(x_i)=0}\frac{\delta(x-x_i)}{|g'(x_i)|}. $$ Despite writing the signed version in a strange way, you apparently used the correct version with absolute value.
Next use $$ \int_a^b f(x)\delta(x-x_i)dx = f(x_i) $$ if $a<x_i<b$.