Question

$\int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx =$

• A. $\frac{1}{2} e^{-2x} \tan x + c$
• B. $-\frac{1}{2} e^{-2x} \tan x + c$
• C. $\frac{1}{2} e^{-2x} \cot x + c$
• D. $-\frac{1}{2} e^{-2x} \cot x + c$

Hint:

Look for integrands that resemble the derivative of a product, especially involving $e^{ax}$ and trigonometric functions.

Answer 1

The Correct Option is A

Step 1: Simplify the trigonometric terms.
Using $1 + \cos 2x = 2 \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$, the integrand becomes:
$$ e^{-2x} \left( \frac{1 - 2 \sin x \cos x}{2 \cos^2 x} \right) = e^{-2x} \left( \frac{1}{2} \sec^2 x - \tan x \right) $$
Step 2: Recognize the form of the derivative of a product.
Consider the derivative of $ \frac{1}{2} e^{-2x} \tan x $: $$ \frac{d}{dx} \left( \frac{1}{2} e^{-2x} \tan x \right) = \frac{1}{2} \left( e^{-2x} (\sec^2 x) + (\tan x) (-2e^{-2x}) \right) $$ $$ = \frac{1}{2} e^{-2x} (\sec^2 x - 2 \tan x) = e^{-2x} \left( \frac{1}{2} \sec^2 x - \tan x \right) $$ This matches the integrand.
Step 3: Integrate both sides.
$$ \int e^{-2x} \left( \frac{1}{2} \sec^2 x - \tan x \right) dx = \int \frac{d}{dx} \left( \frac{1}{2} e^{-2x} \tan x \right) dx $$ $$ I = \frac{1}{2} e^{-2x} \tan x + C $$ Thus, the integral is $ \boxed{\frac{1}{2} e^{-2x} \tan x + c} $.