Answer
a) $y=\tan \theta +C$
b) $y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$
c) $y=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$
Work Step by Step
a) When $\sec^2 \theta=(tan \theta)'$
Thus, $y'=\sec^2 \theta\implies y=\tan \theta +C$
b) When $\sqrt {\theta}=(\dfrac{2}{3} \sqrt{\theta^{3}})'$
Thus, $y'=\sqrt {\theta}\implies y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$
c) The general solution will be the difference of part a and part b, that is, $y'=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$
