Differentiating an accumulation function with a composite upper limit

By the Fundamental Theorem of Calculus Part 1, if $F(x) = \int_a^{g(x)} f(t)\, dt$ where $f$ is continuous, then $F'(x) = f(g(x)) \cdot g'(x)$. Here the integrand is $f(t) = \ln(1+t^2)$ and the upper limit is $g(x) = \cos(x)$, with $g'(x) = -\sin(x)$.

Substituting $g(x) = \cos(x)$ into $f$ gives $f(g(x)) = \ln(1+\cos^2(x))$. Multiplying by $g'(x) = -\sin(x)$ (from the chain rule) gives the final derivative: $F'(x) = \ln(1+\cos^2(x)) \cdot (-\sin(x)) = -\sin(x)\ln(1+\cos^2(x)).$