Answer
$${\bf{i}} - \sqrt 2 {\bf{j}} - \frac{\pi }{4}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {\left( {{{\sec }^2}t{\bf{i}} - 2\cos t{\bf{j}} - {\bf{k}}} \right)} dt \cr
& {\text{use }}\int_a^b {{\bf{r}}\left( t \right)} dt = \left( {\int_a^b {f\left( t \right)} dt} \right){\bf{i}} + \left( {\int_a^b {g\left( t \right)} dt} \right){\bf{j}} + \left( {\int_a^b {h\left( t \right)} dt} \right){\bf{k}}{\text{ }}\left( {{\text{see page 814}}} \right) \cr
& {\text{ then}} \cr
& = \left( {\int_0^{\pi /4} {{{\sec }^2}t} dt} \right){\bf{i}} - \left( {2\int_0^{\pi /4} {\cos t} dt} \right){\bf{j}} - \left( {\int_0^{\pi /4} {dt} } \right){\bf{k}} \cr
& {\text{integrating}} \cr
& = \left( {\tan t} \right)_0^{\pi /4}{\bf{i}} - 2\left( {\sin t} \right)_0^{\pi /4}{\bf{j}} - \left( t \right)_0^{\pi /4}{\bf{k}} \cr
& {\text{evaluating}} \cr
& = \left( {\tan t} \right)_0^{\pi /4}{\bf{i}} - 2\left( {\sin t} \right)_0^{\pi /4}{\bf{j}} - \left( t \right)_0^{\pi /4}{\bf{k}} \cr
& = \left( {\tan \frac{\pi }{4} - \tan 0} \right){\bf{i}} - 2\left( {\sin \frac{\pi }{4} - \sin 0} \right){\bf{j}} - \left( {\frac{\pi }{4} - 0} \right){\bf{k}} \cr
& {\text{simplifying}} \cr
& = \left( {1 - 0} \right){\bf{i}} - 2\left( {\frac{{\sqrt 2 }}{2} - 0} \right){\bf{j}} - \left( {\frac{\pi }{4} - 0} \right){\bf{k}} \cr
& = {\bf{i}} - \sqrt 2 {\bf{j}} - \frac{\pi }{4}{\bf{k}} \cr} $$
