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

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(A) We are given $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.First,simplify the first term: $\frac{2\sin^2 x - 1}{\

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$e^x 	an x + c$

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(A) We are given $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.First,simplify the first term: $\frac{2\sin^2 x - 1}{\cos x} = \frac{-(1 - 2\sin^2 x)}{\cos x} = \frac{-\cos(2x)}{\cos x}$. This does not simplify easily,so let's simplify the whole expression.$f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{-(1 - 2\sin^2 x)}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{-\cos 2x}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.Actually,let's simplify $f(x)$ differently: $f(x) = \frac{2\sin^2 x - (\sin^2 x + \cos^2 x)}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{\sin^2 x - \cos^2 x}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{\sin^2 x}{\cos x} - \cos x + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.Using $1 - \sin^2 x = \cos^2 x$,we have $f(x) = \frac{1 - \cos^2 x}{\cos x} - \cos x + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \sec x - \cos x - \cos x + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \sec x - 2\cos x + \frac{\cos x(2\sin x + 1)}{1 + \sin x}$.This is complex. Let's re-evaluate: $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{2\sin^2 x - 1 + 2\sin^3 x + \sin^2 x}{\cos x} \dots$ No.Let's use $f(x) = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)}{1 + \sin x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x + 1)(1 - \sin x)}{1 - \sin^2 x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{\cos x(2\sin x - 2\sin^2 x + 1 - \sin x)}{\cos^2 x} = \frac{2\sin^2 x - 1}{\cos x} + \frac{-\sin x + 2\sin x + 1 - 2\sin^2 x}{\cos x} = \frac{2\sin^2 x - 1 - \sin x + 2\sin x + 1 - 2\sin^2 x}{\cos x} = \frac{\sin x}{\cos x} = 	an x$.Thus,$f(x) = 	an x$ and $f'(x) = \sec^2 x$.The integral is $\int e^x(	an x + \sec^2 x) dx$.Using the formula $\int e^x(f(x) + f'(x)) dx = e^x f(x) + c$,we get $e^x 	an x + c$.

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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