Question

Evaluate the integral \( \int e^x \left( \frac{1 + \sin x}{1 - \sin x} \right) \, dx \).

Hint:

To solve integrals with trigonometric identities, look for standard transformations (e.g., \( \frac{1 + \sin x}{1 - \sin x} \)) that simplify the integrand.

Answer 1

Step 1: Start by simplifying the given expression. We use the identity for \( \frac{1 + \sin x}{1 - \sin x} \) to convert it into a more convenient form: \[ \frac{1 + \sin x}{1 - \sin x} = \frac{2 \cos^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right)} = 2 \tan \left( \frac{x}{2} \right). \] Step 2: Substitute this identity into the integral: \[ \int e^x \left( \frac{1 + \sin x}{1 - \sin x} \right) dx = \int e^x \cdot 2 \tan \left( \frac{x}{2} \right) dx. \] Step 3: Now, perform substitution. Let: \[ u = \frac{x}{2}, \quad du = \frac{dx}{2}, \quad dx = 2 du. \] The integral becomes: \[ \int e^{2u} \cdot 2 \tan(u) \cdot 2 du = 4 \int e^{2u} \tan(u) du. \] Step 4: Using standard integral techniques, the result of the integral is: \[ e^x \tan \left( \frac{x}{2} \right) + C. \]