# Question #54da5

Jan 16, 2018

$\frac{1}{2} \ln | \sin \left(2 x\right) | + C$

#### Explanation:

$\int \cot \left(2 x\right) \mathrm{dx} = \int \cos \frac{2 x}{\sin} \left(2 x\right) \mathrm{dx}$

you can easily change the numerator to equal the derivative of $\sin \left(2 x\right)$:
$= \frac{1}{2} \int \frac{2 \cos \left(2 x\right)}{\sin} \left(2 x\right) \mathrm{dx} = \frac{1}{2} \int \frac{\frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right)}{\sin} \left(2 x\right) \mathrm{dx}$

now use the rule for integration to $\ln x$:
$\int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + C$

the integral becomes:
$= \frac{1}{2} \cdot \left(\ln | \sin \left(2 x\right) | + C\right)$
$= \frac{1}{2} \ln | \sin \left(2 x\right) | + C$ (1/2 times a constant is still a constant so the C doesn't change)