$\int e^{-2 x} \cos x d x=\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C$ .

Name: $\int e^{-2 x} \cos x d x=\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C$ .
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Question

Quizplus · Accepted Answer

This integral can be solved using integration by parts or the tabular integration method, which involves integrating $e^{-2x}$ and differentiating $\cos x$ repeatedly. The result is indeed $\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C$, indicating the integration has been performed correctly.

∫e−2xcos⁡xdx=15e−2x(−2cos⁡x+sin⁡x)+C\int e^{-2 x} \cos x d x=\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C∫e−2xcosxdx=51​e−2x(−2cosx+sinx)+C

$\int e^{-2 x} \cos x d x=\frac{1}{5} e^{-2 x}(-2 \cos x+\sin x)+C$