Question

The value of \( \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta \) is:

• A. \( e^{\tan \theta} \sec \theta + c \)
• B. \( e^{\tan \theta} \sin \theta + c \)
• C. \( e^{\tan \theta} \sin \theta + c \)
• D. \( e^{\tan \theta} \cos \theta + c \)

Hint:

When dealing with integrals involving trigonometric functions, break the expression into manageable parts and use standard integration identities.

Answer 1

The Correct Option is D

Step 1: We are given: \[ I = \int e^{\tan \theta} (\sec \theta - \sin \theta) d\theta \] Distribute the terms inside the integral: \[ \text{Let } I = \int e^{\tan \theta} (\sec \theta - \sin \theta) \, d\theta \] \[ \text{Put } \tan \theta = t \Rightarrow \sec^2 \theta \, d\theta = dt \Rightarrow d\theta = \frac{dt}{1+t^2} \] \[\Rightarrow I = \int e^t \left(\sqrt{1+t^2} - \frac{t}{\sqrt{1+t^2}}\right) \frac{dt}{1+t^2} \] \[= \int e^t \left(\frac{1}{\sqrt{1+t^2}} - \frac{t}{(1+t^2)^{3/2}}\right) \, dt\] Integrating the first part by parts, we have \[= \frac{1}{\sqrt{1+t^2}} e^t - \int \frac{t}{(1+t^2)^{3/2}} e^t \, dt + \int \frac{t}{(1+t^2)^{3/2}} e^t \, dt + c\] \[= \frac{e^t}{\sqrt{1+t^2}} + c\] \[= e^{\tan \theta} \cos \theta + c\] Thus, the final answer is \( e^{\tan \theta} \cos \theta + c \).