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Find the most general antiderivative of f(x)=−4e^x−1sec^2(x), where −π/2<x<π/2.

Note: Any arbitrary constants used must be an upper-case "C".

since d/dx e^x = e^x

and d/dx tanx = sec^2 x
∫ -4 e^x - sec^2(x) dx = -4e^x - tan(x) + C
note that the domain was conveniently chosen to avoid the asymptotes.

Did you know?

Did you know that finding the most general antiderivative of a function can be quite a challenge? For instance, when faced with the function f(x) = -4e^x - 1sec^2(x), where -π/2 < x < π/2, one must carefully determine its antiderivative. To do this, the antiderivative of each term must be found separately. The antiderivative of -4e^x is simply -4e^x. However, the antiderivative of -1sec^2(x) requires a different approach. By utilizing a known trigonometric identity, the antiderivative of -1sec^2(x) can be found to be equal to -tan(x). To obtain the most general antiderivative of f(x), one must combine both antiderivatives and introduce an arbitrary constant, typically represented by an upper-case "C". Thus, the most general antiderivative of f(x) = -4e^x - 1sec^2(x) is -4e^x - tan(x) + C. Remember, arbitrary constants are essential when working with antiderivatives to account for all infinite possibilities.