Answer
$(\dfrac{\pi+2 \sqrt 2}{2}) j+2k$
Work Step by Step
We need to integrate the given integral as follows:
$[\int_{-\pi/4}^{\pi/4} (\sin t dt)]i+[\int_{-\pi/4}^{\pi/4} (1+\cos t dt)]j+[\int_{-\pi/4}^{\pi/4} (\sec^2 t dt)]k=[-\cos t]_{-\pi/4}^{\pi/4}i+[t+\sin t]_{-\pi/4}^{\pi/4}j+[\tan t]_{-\pi/4}^{\pi/4} k$
or,
$[-\cos t]_{-\pi/4}^{\pi/4}i+[t+\sin t]_{-\pi/4}^{\pi/4}j+[\tan t]_{-\pi/4}^{\pi/4} k=0i+(\dfrac{\pi}{4}+\dfrac{\sqrt 2}{2}) j+2k$
Thus,
$0i+(\dfrac{\pi}{4}+\dfrac{\sqrt 2}{2}) j+2k=(\dfrac{\pi+2 \sqrt 2}{2}) j+2k$
