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