14. \( \int e^{-x} \tan \left(e^{-x}\right) d x= \) (a) \( \ln \left|\cos \left(e^{-x}\right)\right|+c \) (b) \( \ln \left|\sin \left(e^{-x}\right)\right|+c \) (c) \( \ln \left|\tan \left(e^{-x}\right)\right|+c \) (d) \( \ln \left|\cos \left(e^{x}\right)\right|+c \) (e) \( \ln \left|\sin \left(e^{x}\right)\right|+c \)

To solve the integral \(\int e^{-x} \tan \left(e^{-x}\right) d x\), we can use substitution. Let \( u = e^{-x} \), which means that \( du = -e^{-x} dx \) or \( dx = -\frac{du}{u} \). Thus, our integral transforms into: \[ \int e^{-x} \tan \left(e^{-x}\right) d x = -\int \tan(u) \frac{du}{u} \] Next, we know that the derivative of \(\ln |\cos u|\) is \(-\tan(u)\), so integrating \(-\tan(u)\) gives us \(\ln |\cos u| + C\). Substituting back \(u = e^{-x}\), we have: \[ -\int \tan(u) \frac{du}{u} = \ln |\cos(e^{-x})| + C \] Thus, the correct answer is: (a) \( \ln \left|\cos \left(e^{-x}\right)\right| + c \). Now, let's sprinkle in some additional tasty nuggets of knowledge related to this integral! What’s fascinating is that the functions involved (\(\tan\) and \(e^{-x}\)) have beautifully rich histories in mathematics. The tangent function has deep roots stretching back to ancient civilizations and was widely used in astronomy. Its relationship with exponential functions, especially in calculus, showcases the elegance of how different parts of mathematics intertwine. Feeling like making your own mistakes? That’s okay; we learn through them! A common blunder while integrating functions like \(\tan(u)\) is forgetting to adjust the limits of integration when using substitution. Always remember to substitute back to your original variable to avoid losing that connection between the problem and its context!