Let f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} then \int e^x(f(x) + f'(x))dx (where c is the constant of integration)
Step-by-Step Solution
Key Concept: The integral of e^x(f(x) + f'(x)) is e^x f(x) + C. Simplify f(x) first.
f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{-(1 - 2\sin^2 x)}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = -\frac{\cos 2x}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}. Alternatively, simplify f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)(1 - \sin x)}{1 - \sin^2 x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)(1 - \sin x)}{\cos^2 x} = \frac{2\sin^2 x - 1 + 2\sin x - 2\sin^2 x + 1 - \sin x}{\cos x} = \frac{\sin x}{\cos x} = \tan x. Thus, \int e^x(f(x) + f'(x))dx = e^x f(x) + C = e^x \tan x + C.
Correct Answer: A