Resolve integral with importance sample Monte Carlo I'm trying to compute the integral 
$$\int_{a}^{b}(\sin( 1 + x ) + \cos( 1 + x ))e^{-x}\ dx$$
using importance sample Monte Carlo method.
The exercise ask to use Cauchy Distribution to resolve the integral.
Then
$$\frac{1}{n}\sum_{i}^{n}\frac{f(x_i)}{g(x_{i})}$$
is an aproximation to the integral, where $g(x) = \frac{1}{\pi(1+x^2)}$(Cauchy distribution). To generate random numbers in a range $[a, b]$ following Cauchy distribution I used the inverse transformation:
$$X = \tan(\arctan(a) + U(\arctan(b)-\arctan(a)))\quad \Rightarrow\quad \ U \sim(0,1)$$
but I did not get success.
 A: $\newcommand{\+}{^{\dagger}}
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The probability distribution which is normalized in $\ds{\left[a,b\right)}$ is given by:
$$
{\rm P}\pars{x} = {1 \over \arctan\pars{b} - \arctan\pars{a}}\,{1 \over x^{2} + 1}
$$

$$
\int_{a}^{x}{\rm P}\pars{\xi}\,\dd\xi = \int_{0}^{U}\dd U'\quad\imp\quad
{\arctan\pars{X} - \arctan\pars{a} \over \arctan\pars{b} - \arctan\pars{a}}
=U
$$
  which leads to
  $\ds{\arctan\pars{X}
     =\arctan\pars{a} + \bracks{\arctan\pars{b} - \arctan\pars{a}}U}$:
  \begin{align}
X = \tan\pars{\arctan\pars{a} + \bracks{\arctan\pars{b} - \arctan\pars{a}}}
={a + \tan\pars{\bracks{\arctan\pars{b} - \arctan\pars{a}}U}
\over 1 - a\tan\pars{\bracks{\arctan\pars{b} - \arctan\pars{a}}U}}\tag{1}
\end{align}

$$
\sum_{i = 1}^{N}\fermi\pars{x_{i}} \approx
N\int_{a}^{b}{\rm P}\pars{x}\fermi\pars{x}\,\dd x 
$$
$$
\int_{a}^{b}{\fermi\pars{x} \over x^{2} + 1} \approx
{\arctan\pars{b} - \arctan\pars{a} \over N}\sum_{i = 1}^{N}\fermi\pars{x_{i}}
$$

\begin{align}
&\color{#000}{\large%
\int_{a}^{b}\bracks{\sin\pars{1 + x} + \cos\pars{1 + x}}\expo{-x}\,\dd x}
\\[3mm]&\approx\color{#00f}{\large%
{\arctan\pars{b} - \arctan\pars{a} \over N}\times}
\\[3mm]&\color{#00f}{\large\sum_{i = 1}^{N}
\bracks{\sin\pars{1 + x_{i}} + \cos\pars{1 + x_{i}}}\expo{-x_{i}}
\pars{x_{i}^{2} + 1}}
\end{align}
  with $\ds{\braces{x_{i}}}$ generated by $\pars{1}$ and associated $\ds{\braces{U_{i}}}$ generated by an uniform distribution in $\ds{\left[0,1\right)}$.

