# Show that  int \ csc x \ dx = ln|tan(x/2)| + C ?

Jan 4, 2017

We have:

$\int \setminus \csc x \setminus \mathrm{dx} = \int \setminus \csc x \cdot \frac{\csc x - \cot x}{\csc x - \cot x} \setminus \mathrm{dx}$
$\text{ } = \int \setminus \frac{{\csc}^{2} x - \csc x \cot x}{\csc x - \cot x} \setminus \mathrm{dx}$

We now use the following substitution:

$u = \csc x - \cot x \implies \frac{\mathrm{du}}{\mathrm{dx}} = - \csc x \cot x + {\csc}^{2} x$,

And so our integral becomes:

$\int \setminus \csc x \setminus \mathrm{dx} = \int \setminus \frac{1}{u} \setminus \mathrm{du}$
$\text{ } = \ln | u | + C$
$\text{ } = \ln | \csc x - \cot x | + C$

This is the calculus part of the question complete, It now remains to show that this solution is equivalent to the given solution;

We have;

$\csc x - \cot x = \frac{1}{\sin} x - \cos \frac{x}{\sin} x$
$\text{ } = \frac{1 - \cos \left(2 \cdot \frac{1}{2} x\right)}{\sin} \left(2 \cdot \frac{1}{2} x\right)$
$\text{ } = \frac{1 - \left(1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}$
$\text{ } = \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}$
$\text{ } = \frac{\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}$
$\text{ } = \tan \left(\frac{x}{2}\right)$

And so;

$\int \setminus \csc x \setminus \mathrm{dx} = \ln | \csc x - \cot x | + C$
$\text{ "= ln|tan(x/2)| + C " }$, QED