The integral $\int e ^ { 2 x } \cos x d x$ is

A) $\frac { e ^ { x } ( \sin x - 2 \cos x ) } { 5 } + C$
B) $\frac { e ^ { x } ( \sin x + 2 \cos x ) } { 5 } + C$
C) $\frac { e ^ { x } ( 2 \sin x + \cos x ) } { 5 } + C$
D) $\frac { e ^ { x } ( 2 \sin x - \cos x ) } { 5 } + C$
E) $- \frac { e ^ { x } ( \sin x + 2 \cos x ) } { 5 } + C$

Question

The integral $\int e ^ { 2 x } \cos x d x$ is

A) $\frac { e ^ { x } ( \sin x - 2 \cos x ) } { 5 } + C$
B) $\frac { e ^ { x } ( \sin x + 2 \cos x ) } { 5 } + C$
C) $\frac { e ^ { x } ( 2 \sin x + \cos x ) } { 5 } + C$
D) $\frac { e ^ { x } ( 2 \sin x - \cos x ) } { 5 } + C$
E) $- \frac { e ^ { x } ( \sin x + 2 \cos x ) } { 5 } + C$

Quizplus · Accepted Answer

This integral can be solved using integration by parts or by recognizing it as a case for the tabular method, which is a systematic approach for repeated integration by parts. The correct answer involves recognizing that integrating $e^{2x}\cos(x)dx$ will involve both $e^{2x}\sin(x)$ and $e^{2x}\cos(x)$ terms, with coefficients determined by the derivatives and integrals of $e^{2x}$ and $\cos(x)$. The correct result, after simplifying and combining constants, is $\frac { e ^ { x } ( \sin x + 2 \cos x ) } { 5 } + C$, which matches option B.

The Integral ∫e2xcos⁡xdx\int e ^ { 2 x } \cos x d x∫e2xcosxdx

The Integral $\int e ^ { 2 x } \cos x d x$