Antiderivative of a sum of trig derivative forms

Recall that $\frac{d}{dx}[\sec x] = \sec x \tan x$, so an antiderivative of $\sec(x)\tan(x)$ is $\sec x$. Also recall that $\frac{d}{dx}[\cot x] = -\csc^2 x$, so an antiderivative of $\csc^2(x)$ is $-\cot x$.

Since integration is linear, we can antidifferentiate each term separately and add the results. This gives $\sec x - \cot x$, and since $h(x)$ is defined on an interval where these are differentiable, we must add an arbitrary constant $C$ to capture all antiderivatives: $H(x) = \sec x - \cot x + C.$