$\int \frac{2 x sin 2 x}{1 + cos 2 x} d x$ is equal to

x + log | tan x | + C

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x log | tan x | + C

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x tan x + C

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x + tan x + C

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x sec x + C

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Solution

The correct option is C $x tan x + C$
Let $I = \int \frac{2 x sin 2 x}{1 + cos 2 x} d x$
$= \int \frac{2 x}{1 + cos 2 x} + \frac{sin 2 x}{1 + cos 2 x} d x$
$\int \frac{2 x}{2 {cos}^{2} x} + \int tan x d x = \int x {sec}^{2} x + \int tan x d x$
$= x tan x + C$

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