Find the integral.
$$\int e ^ { 2 x } \sin 5 x d x$$
A) $\frac { 1 } { 29 } e ^ { 2 x } ( 2 \sin 5 x - 5 \cos 5 x ) + C$
B) $\frac { 1 } { 29 } e ^ { 2 x } ( 2 \sin 5 x - 5 \cos 2 x ) - C$
C) $\frac { 1 } { 29 } e ^ { 2 x } ( 2 \sin 5 x - 2 \cos 5 x ) + C$
D) $\frac { 1 } { 29 } e ^ { 2 x } ( 2 \sin 5 x - 2 \cos 5 x ) - C$

Question

Quizplus · Accepted Answer

We can apply integration by parts with $u = \sin 5x$ and $dv = e^{2x} dx$ to get
$$\int e^{2x} \sin 5x dx = \frac{1}{2} e^{2x} \sin 5x - \frac{5}{2} \int e^{2x} \cos 5x dx$$
Next, we apply integration by parts again with $u = \cos 5x$ and $dv = e^{2x} dx$:
$$\int e^{2x} \cos 5x dx = \frac{1}{2} e^{2x} \cos 5x + \frac{5}{2} \int e^{2x} \sin 5x dx$$
Substituting this back into the original integral and simplifying, we get
$$\int e^{2x} \sin 5x dx = \frac{1}{29} e^{2x} (2\sin 5x - 5\cos 5x) + C$$
Therefore, the correct answer is (A).

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Find the Integral $\int e ^ { 2 x } \sin 5 x d x$