Answer
$i+\ln (2) j+\dfrac{3}{4}k$
Work Step by Step
We need to integrate the given integral as follows:
$[\int_{0}^{\pi/3} (\sec t \tan t dt)]i+[\int_{0}^{\pi/3} ( \tan t dt)]j+[\int_{0}^{\pi/3} (2 \sin t dt)]k=[\sec t ]_{0}^{\pi/3}i+[-\ln |cos t|]_{0}^{\pi/3} k+[(-1/2) \cos 2 t ]_{0}^{\pi/3} k$
or,
$[\sec t ]_{0}^{\pi/3}i+[-\ln |cos t|]_{0}^{\pi/3} k+[(-1/2) \cos 2 t ]_{0}^{\pi/3} k=(2-1)i+(-\ln \dfrac{-1}{2}-\ln 0) j+\dfrac{3}{4}k$
Thus,
$(2-1)i+(-\ln \dfrac{-1}{2}-\ln 0) j+\dfrac{3}{4}k=i+\ln (2) j+\dfrac{3}{4}k$
