Step 1: The given integral is \( I = \int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta \).
Distribute the terms within the integral:
\[ I = \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta \]
Let \( \tan \theta = t \). Then \( \sec^2 \theta \, d\theta = dt \), which implies \( d\theta = \frac{dt}{1+t^2} \).
\[ \Rightarrow I = \int e^t \left(\sqrt{1+t^2} - \frac{t}{\sqrt{1+t^2}}\right) \frac{dt}{1+t^2} \]
\[= \int e^t \left(\frac{1}{\sqrt{1+t^2}} - \frac{t}{(1+t^2)^{3/2}}\right) \, dt\]
Integrate the first term \( \frac{1}{\sqrt{1+t^2}} e^t \) by parts. This yields:
\[= \frac{1}{\sqrt{1+t^2}} e^t - \int \frac{t}{(1+t^2)^{3/2}} e^t \, dt + \int \frac{t}{(1+t^2)^{3/2}} e^t \, dt + c\]
The integral terms cancel out:
\[= \frac{e^t}{\sqrt{1+t^2}} + c\]
Substitute back \( t = \tan \theta \):
\[= \frac{e^{\tan \theta}}{\sqrt{1+\tan^2 \theta}} + c\]
Since \( \sqrt{1+\tan^2 \theta} = \sec \theta \), we have:
\[= \frac{e^{\tan \theta}}{\sec \theta} + c\]
\[= e^{\tan \theta} \cos \theta + c\]
The final answer is \( e^{\tan \theta} \cos \theta + c \).