Question

$\int\frac{2x+\sin 2x}{1+\cos 2x}dx=$

• A. $x^{2}\sec x+C$
• B. $x+\tan x+C$
• C. $x^{2}\tan x+C$
• D. $x\sec x+C$
• E. $x\tan x+C$

Hint:

Calculus Tip: If differentiating your final answer takes less than 30 seconds, do it! The derivative of $x\tan x$ is $1(\tan x) + x(\sec^2 x)$, which exactly matches our simplified integrand from Step 2.

Answer 1

Answer: (E)

Concept:
The integral can be dramatically simplified by using the double angle identities: $\sin 2x = 2\sin x \cos x$ and $1 + \cos 2x = 2\cos^2 x$. After simplifying, the integral splits into two manageable pieces, one of which requires Integration by Parts.

Step 1: Apply double angle identities.
Substitute the identities into the numerator and denominator: $$I = \int \frac{2x + 2\sin x \cos x}{2\cos^2 x}dx$$

Step 2: Split the fraction and simplify.
Divide both terms in the numerator by the denominator: $$I = \int \left(\frac{2x}{2\cos^2 x} + \frac{2\sin x \cos x}{2\cos^2 x}\right)dx$$ $$I = \int \left(x \sec^2 x + \frac{\sin x}{\cos x}\right)dx$$ $$I = \int (x \sec^2 x + \tan x)dx$$

Step 3: Separate into two integrals.
$$I = \int x \sec^2 x\,dx + \int \tan x\,dx$$

Step 4: Apply Integration by Parts to the first term.
For $\int x \sec^2 x\,dx$, use $\int u\,dv = uv - \int v\,du$. Let $u = x \implies du = dx$. Let $dv = \sec^2 x\,dx \implies v = \tan x$. $$\int x \sec^2 x\,dx = x\tan x - \int \tan x\,dx$$

Step 5: Combine the terms and finalize.
Substitute the evaluated parts back into the total integral expression from
Step 3: $$I = \left(x\tan x - \int \tan x\,dx\right) + \int \tan x\,dx$$ The $\int \tan x\,dx$ terms cancel each other out entirely: $$I = x\tan x + C$$ Hence the correct answer is (E) $x\tan x+C$.