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If an antiderivative of \(f(x)\) is \(\mathrm{e}^{\mathrm{x}}\) and that of \(\mathrm{g}(\mathrm{x})\) is \(\cos x\), then \(\int f(x) \cos x d x+\int g(x) e^{x} d x\) is equal to :
(C) : Given that, \(\int f(x) d x=e^{x}\) \(\text { i.e. } f(x)=e^{x}\) And, \(\quad \int g(x) d x=\cos x\) i.e. \(g(x)=-\sin x\) Then, \(I=\int f(x) \cos x d x+\int g(x) e^{x} d x\) \(I=\int e^{x} \cos x d x+\int-\sin x \cdot e^{x} d x\) \(\mathrm{I}=\cos \mathrm{x} \mathrm{e}^{\mathrm{x}}-\int(-\sin \mathrm{x}) \mathrm{e}^{\mathrm{x}} \mathrm{dx}-\int \sin \mathrm{x} \cdot \mathrm{e}^{\mathrm{x}} \mathrm{dx}\) \(I=e^{x} \cos x+\int \sin x \cdot e^{x} d x-\int \sin x \cdot e^{x} d x\) \(\mathrm{I}=\mathrm{e}^{\mathrm{x}} \cos \mathrm{x}+\mathrm{C}\)
