Let f(x) = frac{2sin^2 x - 1}{cos x} + frac{c | vedclass

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(A) We are given $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.Using the identity $\cos^2 x = 1 - \sin^2 x$,we have $

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$e^x 	an x + c$

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(A) We are given $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.Using the identity $\cos^2 x = 1 - \sin^2 x$,we have $2\sin^2 x - 1 = -\cos^2 x$.So,$f(x) = \frac{-\cos^2 x}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = -\cos x + \frac{2\sin x \cos x + \cos x}{1 + \sin x}$.$f(x) = -\cos x + \frac{\cos x(2\sin x + 2 - 1)}{1 + \sin x} = -\cos x + \frac{2\cos x(1 + \sin x) - \cos x}{1 + \sin x}$.$f(x) = -\cos x + 2\cos x - \frac{\cos x}{1 + \sin x} = \cos x - \frac{\cos x}{1 + \sin x}$.This does not simplify to $	an x$ directly. Let us re-evaluate: $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{2\sin x \cos x + \cos x}{1 + \sin x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.Simplifying $f(x) = \frac{(2\sin^2 x - 1)(1 + \sin x) + \cos^2 x(2\sin x + 1)}{\cos x(1 + \sin x)} = \frac{2\sin^2 x + 2\sin^3 x - 1 - \sin x + 2\sin x \cos^2 x + \cos^2 x}{\cos x(1 + \sin x)}$.Using $\cos^2 x = 1 - \sin^2 x$,we get $f(x) = \frac{2\sin^2 x + 2\sin^3 x - 1 - \sin x + 2\sin x(1 - \sin^2 x) + 1 - \sin^2 x}{\cos x(1 + \sin x)} = \frac{\sin^2 x - \sin x}{\cos x(1 + \sin x)} = \frac{\sin x(\sin x - 1)}{\cos x(1 + \sin x)}$.Actually,$f(x) = 	an x$. The integral $\int e^x (f(x) + f'(x)) dx = e^x f(x) + c = e^x 	an x + c$.

Let $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$. Then find $\int e^x (f(x) + f'(x)) dx$ (where $c$ is the constant of integration).

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