Question

Integrate: \( \int \left(\frac{2 + \sin 2x}{1 + \cos 2x}\right) e^x dx \)

Hint:

When you see an integral involving the product of \( e^x \) and another function, always try to manipulate the other function to see if it can be written as \( f(x) + f'(x) \). This special form is a common shortcut in integration problems.

Answer 1

Step 1: Understanding the Concept:
This integral is in the form \( \int e^x g(x) dx \). This form often simplifies to a standard integral type \( \int e^x (f(x) + f'(x)) dx \), which has a direct solution. The strategy is to simplify the trigonometric function \( g(x) \) and see if it can be expressed as a sum of a function and its derivative.
Step 2: Key Formula or Approach:
The key integration formula is:
\[ \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \] We will also use the trigonometric double-angle identities:
1. \( \sin 2x = 2\sin x \cos x \)
2. \( 1 + \cos 2x = 2\cos^2 x \)
Step 3: Detailed Explanation or Calculation:
Let the integral be \( I \). First, we simplify the trigonometric part of the integrand:
\[ \frac{2 + \sin 2x}{1 + \cos 2x} = \frac{2 + 2\sin x \cos x}{2\cos^2 x} \] Factor out 2 from the numerator and cancel with the denominator:
\[ = \frac{2(1 + \sin x \cos x)}{2\cos^2 x} = \frac{1 + \sin x \cos x}{\cos^2 x} \] Now, split the fraction into two parts:
\[ = \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \] Simplify each part:
\[ = \sec^2 x + \tan x \] Now, substitute this simplified expression back into the integral:
\[ I = \int (\sec^2 x + \tan x) e^x dx \] Let's check if this fits the form \( \int e^x (f(x) + f'(x)) dx \).
If we let \( f(x) = \tan x \), then its derivative is \( f'(x) = \frac{d}{dx}(\tan x) = \sec^2 x \).
So, the integrand is indeed in the form \( e^x (f(x) + f'(x)) \), where \( f(x) = \tan x \).
Using the formula:
\[ I = e^x f(x) + C = e^x \tan x + C \] Step 4: Final Answer:
The result of the integration is \( e^x \tan x + C \).