Given
$$a=\frac{dl}{dc}$$
hence
$$\frac{da}{dc}=\frac{d^2l}{dc^2}$$
Integrate both sides, get
$$\int \frac{da}{dc}dl=\int \frac{d^2l}{dc^2}dl$$
Do a change of variables
$$\int \frac{da}{dc}adc=\int \frac{d^2l}{dc^2}\frac{dl}{dc}dc$$
Now use 1st fundamental theorem of calculus on LHS
$$\int ada=\int \frac{d}{dc}(\frac{dl}{dc})\frac{dl}{dc}dc$$
and then RHS
$$\frac{a^2}{2}+C=\int \frac{dl}{dc}d(\frac{dl}{dc})dc$$
And thus
$$\frac{a^2}{2}+C=\frac{1}{2}(\frac{dl}{dc})^2+C$$
Thus it is valid