Answer
a) $\sec x+C$
b) $\dfrac{4}{3} \sec 3x+C$
c) $\dfrac{2}{\pi} \sec\dfrac{\pi x}{2}+C $
Work Step by Step
a) Since, the anti-derivative for $(\sec x \tan x)$ is: $\sec x+C$
b) The anti-derivative is:
Thus, $[4(\dfrac{1}{3} \sec 3x)]+C=(\dfrac{4}{3}) \sec 3x+C$
c) The anti-derivative is:
Thus, $ ( \dfrac{1}{\pi/2}) \sec (\dfrac{\pi x}{2})+C =(\dfrac{2}{\pi}) \sec (\dfrac{\pi x}{2})+C $