Question

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.

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:

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