Question

The value of \(\int e^x[\frac{1+\sin x}{1+\cos x}]dx\) is equal to

• A. \(e^x\tan\frac{x}{2}+c\)
• B. e^x tanx +c
• C. e^x (1 + cosx) +c
• D. e^x (1 + sinx) +c

Answer 1

The Correct Option is A

The given integral is:

\[ \int e^x \frac{1 + \sin x}{1 + \cos x} \, dx \]

We can simplify the expression inside the integral. By using the identity \( 1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right) \) and \( 1 + \sin x = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \), the expression becomes:

\[ \int e^x \frac{2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)}{2 \cos^2 \left( \frac{x}{2} \right)} \, dx \]

After simplifying:

\[ \int e^x \tan \left( \frac{x}{2} \right) \, dx \]

The integral of \( e^x \tan \left( \frac{x}{2} \right) \) is:

\[ e^x \tan \left( \frac{x}{2} \right) + C \]

Thus, the correct answer is: (A) \( e^x \tan \frac{x}{2} + C \)