Find the indefinite integral below. $$\int \left( 5 e ^ { 5 t } \mathbf { i } - 4 \sin 4 t \mathbf { j } + 6 \cos 3 t \mathbf { k } \right) d t$$ Do not include an arbitrary constant vector.
A) $e ^ { 5 t } \mathbf { i } - \cos 4 t \mathbf { j } + 2 \sin 3 t \mathbf { k }$
B) $e ^ { 5 t } \mathbf { i } + \cos 4 t \mathbf { j } - 2 \sin 3 t \mathbf { k }$
C) $e ^ { 5 t } \mathbf { i } + \cos 4 t \mathbf { j } + 6 \sin 3 t \mathbf { k }$
D) $e ^ { 5 t } \mathbf { i } + \cos 4 t \mathbf { j } + 2 \sin 3 t \mathbf { k }$
E) $\frac { e ^ { 5 t } } { 5 } \mathbf { i } + \cos 4 t \mathbf { j } + 2 \sin 3 t \mathbf { k }$

Question

Quizplus · Accepted Answer

The integral of $5e^{5t}$ is $e^{5t}$ because the derivative of $e^{5t}$ is $5e^{5t}$, so integrating gives back $e^{5t}$. The integral of $-4\sin(4t)$ is $\cos(4t)$ because the derivative of $\cos(4t)$ is $-4\sin(4t)$, and the integral of $6\cos(3t)$ is $2\sin(3t)$ because the derivative of $2\sin(3t)$ is $6\cos(3t)$, taking into account the chain rule for both trigonometric integrals.

Find the Indefinite Integral Below ∫(5e5ti−4sin⁡4tj+6cos⁡3tk)dt\int \left( 5 e ^ { 5 t } \mathbf { i } - 4 \sin 4 t \mathbf { j } + 6 \cos 3 t \mathbf { k } \right) d t∫(5e5ti−4sin4tj+6cos3tk)dt

Find the Indefinite Integral Below $\int \left( 5 e ^ { 5 t } \mathbf { i } - 4 \sin 4 t \mathbf { j } + 6 \cos 3 t \mathbf { k } \right) d t$