Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach $$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$
If $$\tan\left(\frac{x}{2}\right)=u$$
or $$x=2\cdot\tan^{-1}(u)$$
Then,
$$\sin{x}=\dfrac{2u}{1+u^2}$$
$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$
$$dx=\dfrac{2}{1+u^2}$$
Substitute $$\tan\left(\dfrac{x}{2}\right)=u$$
Let us simplify the integrand before integrating
$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$
$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$
$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$
$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$
$$=\dfrac{1+u^2}{4u^2+4u+2}$$
$$=\dfrac{1+u^2}{(2u+1)^2+1}$$
$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$
$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$
Now,
Take :  $$v=2u+1$$
Therefore, $$dv=2\,du$$
$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$
$$I=\tan^{-1}(v)$$
Substitute everything back
$$I=\tan^{-1}(2u+1)$$
$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$
$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$
I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.

With $a=\tan^{-1}\frac1{\sqrt5}$ and $t=\frac12(x+a)$
\begin{align}
\int\frac{dx}{3 + 2\sin x - \cos x}
& = \int \frac{dx}{3-\sqrt5\cos(x+a)}\\
&= \int \frac{2d(\tan^2t)}{(3-\sqrt5)+(3+\sqrt5)\tan^2t}=\tan^{-1} \frac{2\tan t}{3-\sqrt5} +C
\end{align}
