If ∫x+sin⁡x1+cos⁡xdx=f(x)tan⁡x2+C then

A
$f (x) = x^{2}$
B
$f (x) = x$
C
$f (x) = 1 + \cos x$
D
none of these

Solution:

$\int \frac{x + \sin x}{1 + \cos x} d x = \frac{1}{2} \int x \sec^{2} \frac{x}{2} d x + \int \frac{\sin x}{1 + \cos x} d x$

$\begin{matrix} = \frac{1}{2} [2 x \tan \frac{x}{2} - 2 \int \tan \frac{x}{2}] + \int \tan \frac{x}{2} d x \\ = x \tan \frac{x}{2} + C . \end{matrix}$

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If $\int \frac{x + \sin x}{1 + \cos x} d x = f (x) \tan \frac{x}{2} + C$ then

Solution:

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