Let $h(x) = (fog)(x) + K$ where $K$ is any constant. If $\frac{d}{dx}(h(x)) = -\frac{\sin x}{\cos^2(\cos x)}$ if $f(x) = \int_{f(x)}^{g(x)} \frac{f(t)}{g(t)} dt$, where $f$ and $g$ are trigonometric functions then the value of $j(0)$ is equal to $(\cos(1) = .54)$ ____.
Step-by-Step Solution
Key Concept: Recognize that integrating the derivative $-\frac{\sin x}{\cos^2(\cos x)}$ using substitution $u = \cos x$ yields $h(x) = \tan(\cos x) + C$, then identify the composite function components $f(x) = \tan x$ and $g(x) = \cos x$ by comparing with $h(x) = (f \circ g)(x) + K$.
Given $\frac{d}{dx}(h(x)) = -\frac{\sin x}{\cos^2(\cos x)}$, we integrate using substitution $u = \cos x$ to get $h(x) = \tan(\cos x) + C$. Comparing with $h(x) = (fog)(x) + K$, we identify $f(x) = \tan x$ and $g(x) = \cos x$. Then $j(0) = \int_0^{g(0)} \tan t \, dt = \int_0^1 (\tan t \sec t) dt = \sec(1) - 1$.
Correct Answer: 85