Question

$\int \frac{2x + \sin 2x}{1 + \cos 2x}\, dx$ is equal to:

• A. $x + \log|\tan x| + C$
• B. $x \log|\tan x| + C$
• C. $x \tan x + C$
• D. $x + \tan x + C$
• E. $x \sec x + C$

Hint:

Whenever you see an integral involving $(1 + \cos 2x)$ in the denominator, your first move should almost always be to convert it to $2 \cos^2 x$. This usually leads to a much simpler expression involving $\sec^2 x$.

Answer 1

The Correct Option is C

Concept: To solve this integral, we simplify the integrand using double-angle trigonometric identities.
• $1 + \cos 2x = 2 \cos^2 x$
• $\sin 2x = 2 \sin x \cos x$

Step 1: Simplify the integrand using identities.
Substitute the identities into the integral: \[ \int \frac{2x + 2 \sin x \cos x}{2 \cos^2 x} \, dx \] Divide each term in the numerator by the denominator: \[ \int \left( \frac{2x}{2 \cos^2 x} + \frac{2 \sin x \cos x}{2 \cos^2 x} \right) \, dx \] \[ = \int (x \sec^2 x + \tan x) \, dx \]

Step 2: Recognize the integral form.
The resulting expression is in the form $\int [f(x) + x f'(x)] \, dx$, which evaluates to $x f(x) + C$.
• Let $f(x) = \tan x$
• Then $f'(x) = \sec^2 x$ So, $\int (x \sec^2 x + \tan x) \, dx = x \tan x + C$.