## University Calculus: Early Transcendentals (3rd Edition)

$i-j+\dfrac{\pi}{4}k$
Plug $p= 2t \implies dp=dt$ and $\cos 2t =1-2 \sin^2 t$ We need to integrate the given integral as follows: $[\int_{0}^{\pi/2} \cos t dt]i-[\int_{0}^{\pi/2} \sin 2t dt]j+[\int_{0}^{\pi/2} \sin^2 t dt]k=[\int_{0}^{\pi/2} (1-2 \sin^2 t) dt]i-[\int_{0}^{\pi/2} \sin p dp]j+[\int_{0}^{\pi/2} (\dfrac{1}{2}-\dfrac{1}{2} \cos 2t dt]k$ or, $[\int_{0}^{\pi/2} (1-2 \sin^2 t) dt]i-[\int_{0}^{\pi/2} \sin p dp]j+[\int_{0}^{\pi/2} (\dfrac{1}{2}-\dfrac{1}{2} \cos 2t dt]k=[\sin t]_{0}^{\pi/2}i+[\dfrac{1}{2} \cos 2t ]_{0}^{\pi/2}j+[(\dfrac{1}{2}t-\dfrac{1}{4} \sin 2t]_{0}^{\pi/2}$ Thus, $[\sin t]_{0}^{\pi/2}i+[\dfrac{1}{2} \cos 2t ]_{0}^{\pi/2}j+[(\dfrac{1}{2}t-\dfrac{1}{4} \sin 2t]_{0}^{\pi/2}=i-j+\dfrac{\pi}{4}k$