Question

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

• A. $x \tan\left(\frac{x}{2}\right) + c$
• B. $\log(x + \sin x) + c$
• C. (x2) + c
• D. $\log(1 + \cos x) + c$

Hint:

This problem perfectly follows the integration template $\int [f(x) + xf'(x)]dx = xf(x) + c$. If you rewrite the expression, it reveals itself as $\int [\tan(x/2) + x \cdot \frac{1}{2}\sec^2(x/2)]dx$. Recognizing this template allows you to instantly write down the answer $x\tan(x/2)$ without doing any integration steps!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem presents an indefinite trigonometric integral. We need to simplify the fractional integrand expression using half-angle identities to find its antiderivative.

Step 2: Key Formula or Approach:
We use the standard trigonometric half-angle identities to replace the terms in both the numerator and denominator: $$ 1 + \cos x = 2\cos^2\left(\frac{x}{2}\right) $$ $$ \sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) $$

Step 3: Detailed Explanation:
Let's rewrite the integral $I$ by substituting these half-angle formulas into the denominator: $$ I = \int \frac{x + \sin x}{2\cos^2\left(\frac{x}{2}\right)} dx $$ Now, split the integral into two separate algebraic fractions: $$ I = \int \frac{x}{2\cos^2\left(\frac{x}{2}\right)} dx + \int \frac{\sin x}{2\cos^2\left(\frac{x}{2}\right)} dx $$ Using $\frac{1}{\cos^2\theta} = \sec^2\theta$ for the first part, and replacing $\sin x$ with its double-angle identity in the second part: $$ I = \frac{1}{2} \int x \sec^2\left(\frac{x}{2}\right) dx + \int \frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\cos^2\left(\frac{x}{2}\right)} dx $$ Simplify the second integral by canceling out common terms: $$ I = \frac{1}{2} \int x \sec^2\left(\frac{x}{2}\right) dx + \int \tan\left(\frac{x}{2}\right) dx $$ Let's evaluate the first integral using Integration by Parts ($\int u dv = uv - \int v du$), setting $u = x$ and $dv = \sec^2\left(\frac{x}{2}\right)dx$:

• $du = dx$

• $v = \int \sec^2\left(\frac{x}{2}\right) dx = 2\tan\left(\frac{x}{2}\right)$
Applying the parts formula to that specific component: $$ \frac{1}{2} \left[ x \cdot 2\tan\left(\frac{x}{2}\right) - \int 2\tan\left(\frac{x}{2}\right) dx \right] = x\tan\left(\frac{x}{2}\right) - \int \tan\left(\frac{x}{2}\right) dx $$ Now, substitute this expanded part back into our full equation for $I$: $$ I = \left( x\tan\left(\frac{x}{2}\right) - \int \tan\left(\frac{x}{2}\right) dx \right) + \int \tan\left(\frac{x}{2}\right) dx $$ Notice that the remaining integral terms cancel out perfectly: $$ I = x\tan\left(\frac{x}{2}\right) + c $$

Step 4: Final Answer:
The value of the indefinite integral is $x \tan\left(\frac{x}{2}\right) + c$, which corresponds to option (A).