Question

Compute the indefinite integral of the following functions. $$\mathbf{r}(t)=t e^{t} \mathbf{i}+t \sin t^{2} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k}$$

Short answer

Answer: The indefinite integral of the vector function \(\mathbf{r}(t)\) is: \(\int \mathbf{r}(t) dt= (te^t - e^t + C_1) \mathbf{i} + (-\frac{1}{2}\cos(t^2)+C_2) \mathbf{j} + (-2\sqrt{t^2+4} + C_3) \mathbf{k}\)

Step-by-step solution

Integrate the first function, \(te^t\)

To integrate \(te^t\), we will use integration by parts, which states: $$\int u dv = uv- \int v du$$ Let \(u=t\) and \(dv=e^t dt\). Then, we have: \(du=dt\) and \(v=e^t\). Now, apply the integration by parts formula: $$\int te^t dt = te^t - \int e^t dt$$ Next, integrate \(e^t\): $$\int e^t dt = e^t + C_1$$ Finally, find the result: $$\int te^t dt = te^t - e^t + C_1$$

Integrate the second function, \(t\sin(t^2)\)

To integrate \(t\sin(t^2)\), we will use the substitution method. Let \(u=t^2\) and \(du=2t\,dt\). Then, the integral becomes: $$\int t\sin(t^2) dt = \frac{1}{2}\int \sin(u) du$$ Now, integrate \(\sin(u)\): $$\int \sin(u) du = -\cos(u)+C_2$$ Substitute \(u=t^2\) back in and find the result: $$\int t\sin(t^2) dt = -\frac{1}{2}\cos(t^2)+C_2$$

Integrate the third function, \(-\frac{2t}{\sqrt{t^2+4}}\)

To integrate \(-\frac{2t}{\sqrt{t^2+4}}\), we will use the substitution method again. Let \(u=t^2+4\) and \(du=2t\,dt\). Then, the integral becomes: $$-\int \frac{2t}{\sqrt{t^2+4}} dt=-\int \frac{1}{\sqrt{u}} du$$ Now, integrate \(\frac{1}{\sqrt{u}}\): $$-\int \frac{1}{\sqrt{u}} du = -2\sqrt{u} +C_3$$ Substitute \(u=t^2+4\) back in and find the result: $$\int -\frac{2t}{\sqrt{t^2+4}} dt = -2\sqrt{t^2+4} + C_3$$

Combine the results to form the indefinite integral vector function

Now that we have integrated all three functions, we will combine them to create the indefinite integral vector function: $$\int \mathbf{r}(t) dt= (te^t - e^t + C_1) \mathbf{i} + (-\frac{1}{2}\cos(t^2)+C_2) \mathbf{j} + (-2\sqrt{t^2+4} + C_3) \mathbf{k}$$ So, the indefinite integral of \(\mathbf{r}(t)\) is: $$\int \mathbf{r}(t) dt= (te^t - e^t + C_1) \mathbf{i} + (-\frac{1}{2}\cos(t^2)+C_2) \mathbf{j} + (-2\sqrt{t^2+4} + C_3) \mathbf{k}$$