for r(t)=t^{1}e^{t^{2}}mathbf{i}+cos(6t)sin(6t)mathbf{j}, find int r(t)dt = mathbf{i}mathbf{i}+mathbf{j}+c…

for r(t)=t^{1}e^{t^{2}}mathbf{i}+cos(6t)sin(6t)mathbf{j}, find int r(t)dt = mathbf{i}mathbf{i}+mathbf{j}+c with constant vector c.

for r(t)=t^{1}e^{t^{2}}mathbf{i}+cos(6t)sin(6t)mathbf{j}, find int r(t)dt = mathbf{i}mathbf{i}+mathbf{j}+c with constant vector c.

Answer

Explanation:

Step1: Integrate the i - component

We know that if (r(t)=t e^{t^{2}}\mathbf{i}+\cos(6t)\sin(6t)\mathbf{j}), for the (\mathbf{i}) - component, let (u = t^{2}), then (du=2tdt) and (\int t e^{t^{2}}dt=\frac{1}{2}\int e^{u}du=\frac{1}{2}e^{t^{2}}+C_1).

Step2: Integrate the j - component

For the (\mathbf{j}) - component, use the double - angle formula (\cos(6t)\sin(6t)=\frac{1}{2}\sin(12t)). Then (\int\cos(6t)\sin(6t)dt=\int\frac{1}{2}\sin(12t)dt). Let (v = 12t), (dv = 12dt), so (\int\frac{1}{2}\sin(12t)dt=-\frac{1}{24}\cos(12t)+C_2).

Answer:

(\frac{1}{2}e^{t^{2}}\mathbf{i}-\frac{1}{24}\cos(12t)\mathbf{j}+\mathbf{C})