Example 2. Calculate $\int_{0}^{\pi / 2} \mathbf{a}(t) d t$, where $\mathbf{a}(t)=\mathbf{i} \cos t-\mathbf{j} \sin ^{2} t$.
Solution. By formula (3)

$\int_{0}^{\pi / 2} \mathbf{a}(t) d t=\mathbf{i} \int_{0}^{\pi / 2} \cos t d t-\mathbf{j} \int_{0}^{\pi / 2} \sin ^{2} t d t=\left.\mathbf{i} \sin t\right|_{0} ^{\pi / 2}-\left.\mathbf{j}\left(\frac{t}{2}-\frac{\sin 2 t}{4}\right)\right|_{0} ^{\pi / 2}=\mathbf{i}-\frac{\pi}{4} \mathbf{j} . \quad$

## Problems for Independent Solution

Calculate the following integrals:

32. $\int_{0}^{\pi} \mathrm{a}(t) d t$, where $\mathrm{a}=\sin ^{2} t \cos t \mathrm{i}+\cos ^{2} t \sin t \mathrm{j}+\mathrm{k}$.
33. $\int_{0}^{1} \mathbf{a}(t) d t$, where $\mathbf{a}=\frac{e^{-i / 2}}{2} \mathbf{i}+\frac{e^{i / 2}}{2} \mathbf{j}+e^{t} \mathbf{k}$.
34. $\int_{0}^{1} \mathbf{a}(t) d t$, where $\mathbf{a}=3 \pi \cos \pi t \mathrm{i}-\frac{1}{1+t} \mathbf{j}+2 t \mathbf{k}$.
35. $\int_{0}^{1} \mathbf{a}(t) d t$, where $\mathbf{a}=(2 t+\pi) \mathbf{i}+t \sin t \mathbf{j}+\pi \mathbf{k}$.