Answer
$$
\int 2^{\tan \theta} \sec ^{2} \theta d \theta =\frac{2^{\tan \theta}}{\ln 2}+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int 2^{\tan \theta} \sec ^{2} \theta d \theta
$$
$$\text { Let } u=\tan \theta . \text { Then } d u=\sec ^{2} \theta d \theta
$$
substituting in the given integral we have :
$$
\begin{aligned}
\int 2^{\tan \theta} \sec ^{2} \theta d \theta & =\int 2^{u} d u
\\
&=\frac{2^{u}}{\ln 2}+C \\
&=\frac{2^{\tan \theta}}{\ln 2}+C
\end{aligned}
$$
where $C$ is an arbitrary constant.