Answer
a) $\ln t +C$
b) $\tan\theta+C$
Work Step by Step
Part a)
We know that $\frac{d}{dt}(\ln t)=\frac{1}{t}$.
Then,
the antiderivative of $g(t)=\frac{1}{t}$ is $G(t)=\ln t+C$.
Part b)
We know that $\frac{d}{d\theta}(\tan\theta)=\sec^2\theta$.
Then,
the antiderivative of $r(\theta)=\sec^2\theta$ is $R(\theta)=\tan\theta+C$.
