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.

Remember the following trigonometric equations, which are always very helpful to solve integrals:
$$
\begin{matrix}
2\cos(u)^2 = 1 + \cos(2u) & & & \left(\cos(u) + \sin(u)\right)^2 = 1 + \sin(2u) \\
2\sin(u)^2 = 1 - \cos(2u) & & & \left(\cos(u) - \sin(u)\right)^2 = 1 - \sin(2u)  \\
\end{matrix}
$$
Note that you can make a mnemonic to remember these equations.

With them, you can solve the integral:
$$
I = \int \dfrac{dx}{3+2\sin(x)-\cos(x)}
$$
Using the following procedure:


*

*Replace with $\;\;u=\frac{x}{2} \;\;\Rightarrow\;\; dx = 2 du$:
$$
\begin{align}
I &= 2 \int \dfrac{du}{3+2\sin(2u)-\cos(2u)}
\end{align}
$$

*Rewrite the expression to a known trigonometric expression:
$$
\begin{align}
I&= 2 \int \dfrac{du}{2\left(1+\sin(2u)\right) + \left(1-\cos(2u)\right)}
\end{align}
$$

*Replace with the corresponding trigonometric equations:
$$
\begin{align}
I&= 2 \int \dfrac{du}{2\left(\cos(u) + \sin(u)\right)^2 + 2\sin(u)^2}
\end{align}
$$


*Extract common factor by non-binomial term:
$$
\begin{align}
I&= 2 \int \dfrac{1}{\left(\dfrac{\cos(u)}{\sin(u)} + 1\right)^2 + 1 } \cdot\dfrac{1}{2\sin(u)^2} du = \int \dfrac{1}{\left(\cot(u) + 1\right)^2 + 1 } \cdot \csc(u)^2 du
\end{align}
$$


*Replace with $\;\;\omega=\cot(u) + 1 \;\;\Rightarrow\;\; d\omega = -\csc(u)^2 du$:
$$
\begin{align}
I&= \int \dfrac{-d\omega}{\omega^2 + 1 } = \mbox{arccot}(\omega) + C
\end{align}
$$

*Returning to the variable x:
$$
\omega =\cot(u) + 1 = \cot\left(\dfrac{x}{2}\right) + 1 \quad\Rightarrow\quad I = \mbox{arccot}\left(\cot\left(\dfrac{x}{2}\right) + 1\right) + C
$$

With a little more effort and using the same procedure, you could solve the following generic integral:
$$
I = \int \dfrac{dx}{A+B\sin(x)+C\cos(x)}
$$
The secret to this is:


*

*In Step 2: choose the correct "known trigonometric expression" according to the signs of sine and cosine. 

*In Step 4: don't forget to remove the non-binomial term to reduce it to a simple expression.
