Step 1: Understanding the Concept:
The integral contains both algebraic ($x$) and trigonometric ($\sin x$, $\cos x$) terms. A common strategy is to use half-angle formulas to simplify the denominator and split the fraction into manageable parts, often leading to integration by parts.
Step 2: Key Formula or Approach:
Use trigonometric half-angle identities:
$1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$
$\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$
Integration by parts: $\int u dv = uv - \int v du$
Step 3: Detailed Explanation:
Let $I = \int \frac{x + \sin x}{1 + \cos x} dx$
Apply half-angle identities to the numerator and denominator:
\[ I = \int \frac{x + 2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} dx \]
Separate the fraction into two integrals:
\[ I = \int \left( \frac{x}{2\cos^2(x/2)} + \frac{2\sin(x/2)\cos(x/2)}{2\cos^2(x/2)} \right) dx \]
Simplify each term:
\[ I = \int \frac{1}{2} x \sec^2\left(\frac{x}{2}\right) dx + \int \frac{\sin(x/2)}{\cos(x/2)} dx \]
\[ I = \int \frac{1}{2} x \sec^2\left(\frac{x}{2}\right) dx + \int \tan\left(\frac{x}{2}\right) dx \]
Now, apply integration by parts to the first integral. Let $u = x$ and $dv = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) dx$.
Then $du = dx$. To find $v$, integrate $dv$:
\[ v = \int \frac{1}{2} \sec^2\left(\frac{x}{2}\right) dx = \frac{1}{2} \cdot \frac{\tan(x/2)}{1/2} = \tan\left(\frac{x}{2}\right) \]
Using the integration by parts formula $\int u dv = uv - \int v du$:
\[ \int \frac{1}{2} x \sec^2\left(\frac{x}{2}\right) dx = x \tan\left(\frac{x}{2}\right) - \int \tan\left(\frac{x}{2}\right) dx \]
Substitute this back into the expression 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 integrals of $\tan(x/2)$ cancel each other out perfectly:
\[ I = x \tan\left(\frac{x}{2}\right) + c \]
Step 4: Final Answer:
The value of the integral is $x \tan \frac{x}{2} + c$.