3. ∫₀^(π/2) (cos 2t)i+(sin 3t)j dt

Explanation:

Step1: Integrate the i - component

We know that $\int\cos(2t)dt=\frac{1}{2}\sin(2t)+C$. Evaluating $\int_{0}^{\frac{\pi}{2}}\cos(2t)dt=\left[\frac{1}{2}\sin(2t)\right]_{0}^{\frac{\pi}{2}}$. Substitute the upper and lower limits: $\frac{1}{2}\sin(2\times\frac{\pi}{2})-\frac{1}{2}\sin(2\times0)=\frac{1}{2}\sin(\pi)-\frac{1}{2}\sin(0)=0 - 0=0$.

Step2: Integrate the j - component

We know that $\int\sin(3t)dt=-\frac{1}{3}\cos(3t)+C$. Evaluating $\int_{0}^{\frac{\pi}{2}}\sin(3t)dt=\left[-\frac{1}{3}\cos(3t)\right]_{0}^{\frac{\pi}{2}}$. Substitute the upper and lower limits: $-\frac{1}{3}\cos(3\times\frac{\pi}{2})+\frac{1}{3}\cos(3\times0)=-\frac{1}{3}\cos(\frac{3\pi}{2})+\frac{1}{3}\cos(0)=0+\frac{1}{3}=\frac{1}{3}$.

Answer:

$\frac{1}{3}\mathbf{j}$