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

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$\frac{θ \sin^{2} θ}{\cos θ} + c$

$θ \cos^{2} θ + c$

$\frac{θ T a n θ}{\sec^{2} θ} + c$

$\frac{θ \sin θ}{\cos θ} + c$

answer is D.

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Detailed Solution

$\begin{array}{l} \int \frac{2 θ + \sin 2 θ}{1 + \cos 2 θ} d θ = \int (θ \sec^{2} θ + T a n θ) d θ \\ = θ T a n θ + c \end{array}$

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