Answer
a) $y=-\dfrac{\cos 2t}{2} +C$
b) $y=2 \sin \dfrac{t}{2} +C$
c) $y=-\dfrac{\cos 2t}{2} +2 \sin \dfrac{t}{2}+C$
Work Step by Step
a) When $\sin 2t=(-\dfrac{\cos 2t}{2})'$
Thus, $y'= \sin 2t\implies y=-\dfrac{\cos 2t}{2} +C$
b) When $\cos \dfrac{t}{2}=( 2 \sin \dfrac{t}{2})'$
Thus, $y'= \cos \dfrac{t}{2} \implies y=2 \sin \dfrac{t}{2} +C$
c) Here, $y'=\sin 2t+\cos \dfrac{t}{2}$
Thus, $y'=-\dfrac{\cos 2t}{2} +2 \sin \dfrac{t}{2}+C$
