If $\int e^{2 x} (\cos x + 7 \sin x) d x = e^{2 x} g (x) + C$ where $c$ is a constant of integration, then $g (0) + g (\frac{π}{2})$ is equal to

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Detailed Solution

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$\begin{array}{l} \int e^{a x} \sin b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x) + C \\ \int e^{a x} \cos b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \cos b x + b \sin b x) + C \\ \int e^{2 x} (\cos x + 7 \sin x) d x \\ = \int e^{2 x} \cos x d x + 7 \int e^{2 x} \sin x d x \\ = \frac{e^{2 x}}{5} (2 \cos x + \sin x) + \frac{7}{5} e^{2 x} (2 \sin x - \cos x) + C \\ = \frac{e^{2 x}}{5} (- 5 \cos x + 15 \sin x) + C \\ \begin{array}{ll} ∴ g (x) = \frac{1}{5} (- 5 \cos x + 15 \sin x) = (- \cos x + 3 \sin x) \\ \Rightarrow g (0) + g (\frac{π}{2}) = - 1 + 3 = 2 \end{array} \end{array}$