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