# How do you integrate int (csc2x)dx?

Jun 5, 2018

$I = \frac{1}{2} \ln | \csc 2 x - \cot 2 x | + c$

#### Explanation:

We know that,

$\textcolor{red}{\int \csc \theta d \theta = \ln | \csc \theta - \cot \theta | + c} \ldots \ldots \ldots . \left(1\right)$

Here,

$I = \int \left(\csc 2 x\right) \mathrm{dx}$

Let, $2 x = u \implies 2 \mathrm{dx} = \mathrm{du} \implies \mathrm{dx} = \frac{1}{2} \mathrm{du}$

$I = \int \csc u \cdot \frac{1}{2} \mathrm{du}$

=1/2intcscudu...tocolor(red)(Apply(1)

$= \frac{1}{2} \ln | \csc u - \cot u | + c , w h e r e , u = 2 x$

$= \frac{1}{2} \ln | \csc 2 x - \cot 2 x | + c$

Jun 5, 2018

$\int \csc 2 x \cdot \mathrm{dx} = \frac{1}{2} \ln \left(\tan x\right) + C$

#### Explanation:

$\int \csc 2 x \cdot \mathrm{dx}$

=$\int \frac{\mathrm{dx}}{\sin 2 x}$

=$\int \frac{\mathrm{dx}}{2 \sin x \cdot \cos x}$
=$\int \frac{\left[{\left(\sin x\right)}^{2} + {\left(\cos x\right)}^{2}\right] \cdot \mathrm{dx}}{2 \sin x \cdot \cos x}$

=$\frac{1}{2} \int \frac{\sin x \cdot \mathrm{dx}}{\cos} x + \frac{1}{2} \int \frac{\cos x \cdot \mathrm{dx}}{\sin} x$

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

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

=$\frac{1}{2} \ln \left(\tan x\right) + C$