Answer
$$\mathbf{r}(t)=4 \cos t \mathbf{j}+3 \sin t \mathbf{k}$$
Work Step by Step
Given$$ \mathbf{r}^{\prime \prime}(t)=-4 \cos t \mathbf{j}-3 \sin t \mathbf{k}, \quad \mathbf{r}^{\prime}(0)=3 \mathbf{k}, \quad \mathbf{r}(0)=4 \mathbf{j}$$
As we have $$\mathbf{r^{\prime}}(t)=\int \mathbf{r}^{\prime \prime}(t) dt$$
and
$$\mathbf{r^{}}(t)=\int \mathbf{r}^{\prime }(t) dt$$
So, integrating on a component-by-component basis produces
\begin{array}{l}
{\mathbf{r^{\prime}}(t)=\int \mathbf{r}^{ \prime\prime}(t)(t)dt=\int\left(-4 \cos t \mathbf{j}-3 \sin t \mathbf{k}\right) d t\\
= -4 \sin t \mathbf{j}+3 \cos t \mathbf{k}+\mathbf{C_1}} \\
\text{since} \ \mathbf{r^{\prime}}(0)=3 \mathbf{k}, \quad \text{we get}
\\ {\mathbf{r^{\prime}}(0)= -4 \sin 0 \mathbf{j}+3\cos 0 \mathbf{k}+\mathbf{C_1}=3 \mathbf{k}}
\\ { \Rightarrow \mathbf{C_1}=0} \\
{\text {so we get }}\\
{\mathbf{r^\prime}(t)=-4 \sin t \mathbf{j}+3 \cos t \mathbf{k}}\\
\\
{\text{Also we have}}\\
{\mathbf{r}(t)=\int \mathbf{r}^{\prime}(t)(t)dt=\int\left(-4 \sin t \mathbf{j}+3 \cos t \mathbf{k}\right) d t\\
= 4 \cos t \mathbf{j}+3 \sin t \mathbf{k}+\mathbf{C}} \\
\text{since} \ \mathbf{r}(0)=4 \mathbf{j}, \quad \text{so we get}
\\ {\mathbf{r}(0)= 4 \cos 0 \mathbf{j}+3\sin 0\mathbf{k}+\mathbf{C}=4 \mathbf{j}}
\\ { \Rightarrow \mathbf{C}=0} \\
{\text {so we get }}\\
{\mathbf{r}(t)=4 \cos t \mathbf{j}+3 \sin t \mathbf{k}}
\end{array}
