# The integral value=?

Mar 15, 2018

$2 \ln \left(\sin \left(\frac{x}{2}\right)\right) + C$

#### Explanation:

$1 + 2 \cot x \left(\csc x + \cot x\right) = 1 + 2 \cos \frac{x}{\sin} x \left(\frac{1}{\sin} x + \cos \frac{x}{\sin} x\right)$
$q \quad = 1 + 2 \frac{\cos x \left(1 + \cos x\right)}{\sin} ^ 2 x = \frac{{\sin}^{2} x + 2 \cos x + 2 {\cos}^{2} x}{\sin} ^ 2 x$
$q \quad = \frac{1 + 2 \cos x + {\cos}^{2} x}{\sin} ^ 2 x = {\left(\frac{1 + \cos x}{\sin} x\right)}^{2}$

So, in the interval $x \in \left(0 , \frac{\pi}{2}\right)$, we have :

$\sqrt{1 + 2 \cot x \left(\csc x + \cot x\right)} = \frac{1 + \cos x}{\sin} x = \frac{2 {\cos}^{2} \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} = \cot \left(\frac{x}{2}\right)$

Thus, the integral is

$\int \sqrt{1 + 2 \cot x \left(\csc x + \cot x\right)} \mathrm{dx} = \int \cot \left(\frac{x}{2}\right) \mathrm{dx} = 2 \int \cot \left(\frac{x}{2}\right) d \left(\frac{x}{2}\right) = 2 \ln \left(\sin \left(\frac{x}{2}\right)\right) + C$