Question

find the most general antiderivative of ( f(x) = -5e^x - 9sec^2(x) ), where ( -\frac{pi}{2} < x < \frac{pi}{2} ).
note: any arbitrary constants used must be an upper-case \c\.
( f(x) = square )

Explanation:

Step1: Antiderivative of \(-5e^x\)

The antiderivative of \(e^x\) is \(e^x\) (by the rule \(\int e^x dx = e^x + C\)). So for \(-5e^x\), the antiderivative is \(-5e^x\) (since the constant factor -5 can be factored out of the integral: \(\int -5e^x dx = -5\int e^x dx = -5e^x + C_1\), where \(C_1\) is a constant).

Step2: Antiderivative of \(-9\sec^2(x)\)

The antiderivative of \(\sec^2(x)\) is \(\tan(x)\) (by the rule \(\int \sec^2(x) dx = \tan(x) + C\)). So for \(-9\sec^2(x)\), the antiderivative is \(-9\tan(x)\) (factoring out the constant -9: \(\int -9\sec^2(x) dx = -9\int \sec^2(x) dx = -9\tan(x) + C_2\), where \(C_2\) is a constant).

Step3: Combine and Add Constant

Combining the two antiderivatives and combining the constants \(C_1\) and \(C_2\) into a single upper - case constant \(C\) (since the sum of two constants is also a constant), we get the antiderivative \(F(x)\) of \(f(x)=-5e^x - 9\sec^2(x)\) as \(F(x)=-5e^x-9\tan(x)+C\).

Answer:

\(-5e^x - 9\tan(x)+C\)