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If an anti-derivative of f(x) is e^x and that of g(x) is cos x, then $\int f (x) \cos xdx + \int g (x) e^{x} dx$ is equal to

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f (x) \cdot g (x) + C

f (x) + g (x) + C

e^{x} \cos x + C

f (x) - g (x) + 1

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detailed solution

Correct option is C

$\int f (x) \cos xdx + \int g (x) e^{x} dx$

$\begin{array}{l} = \frac{e^{x}}{2} (\cos x + \sin x) - \frac{e^{x}}{2} (\sin x - \cos x) + C \\ = \frac{e^{x}}{2} (2 \cos x) + C = e^{x} \cos x + C \end{array}$

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If an anti-derivative of f(x) is ex and that of g(x) is cos x, then ∫f(x)cos ⁡xdx+∫g(x)exdx is equal to