If an antiderivative of f(x) is e^x and an an | vedclass

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(C) Given that the antiderivative of $f(x)$ is $e^x$, we have $\int f(x) \, dx = e^x$, which implies $f(x) = \frac{d}{dx}(e^x) = e^x$.Given that the a

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$e^x \cos x + c$

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(C) Given that the antiderivative of $f(x)$ is $e^x$, we have $\int f(x) \, dx = e^x$, which implies $f(x) = \frac{d}{dx}(e^x) = e^x$.Given that the antiderivative of $g(x)$ is $\cos x$, we have $\int g(x) \, dx = \cos x$, which implies $g(x) = \frac{d}{dx}(\cos x) = -\sin x$.We need to evaluate $I = \int f(x) \cos x \, dx + \int g(x) e^x \, dx$.Substituting the expressions for $f(x)$ and $g(x)$:$I = \int e^x \cos x \, dx + \int (-\sin x) e^x \, dx$.$I = \int e^x \cos x \, dx - \int e^x \sin x \, dx$.Using the product rule for differentiation: $\frac{d}{dx}(e^x \cos x) = e^x \cos x + e^x(-\sin x) = e^x \cos x - e^x \sin x$.Integrating both sides with respect to $x$:$e^x \cos x = \int (e^x \cos x - e^x \sin x) \, dx + c$.$e^x \cos x = \int e^x \cos x \, dx - \int e^x \sin x \, dx + c$.Thus, $I = e^x \cos x + c$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

If an antiderivative of $f(x)$ is $e^x$ and an antiderivative of $g(x)$ is $\cos x$, then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx =$

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