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Q)

# Integrate : \begin{align*} \int \frac{dx}{3+2 \sin x + \cos x} \end{align*}

$(A)\; 1 + \tan^{-1} \;x/2 +C$
$(B) \; \tan^{-1} (1+ \tan^{-1} (x/2)) +C$
$(C) \; \tan^{-1} (1- \tan^{-1} (x/2)) +C$
$(D)\; 1 - \tan^{-1} \;x/2 +C$

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A)
Put $\sin x = \frac{2 \tan x/2}{1 + \tan ^2 x/2}$ and $\cos x = \frac{1-\tan^2 x/2}{1+ \tan^2 x/2}$
\therefore I = \begin{align*}\int \frac{dx}{ 3+ 2 . (\frac{2 \tan x/2}{1+tan^2 x/2})+ (\frac{1-\tan^2 x/2}{1+\tan^2 x/2} )} \end{align*}
= \begin{align*}\int \frac{ (1+ \tan^2 x/2) dx}{3 +3 \tan^2 x/2 + 4 \tan x/2 + 1-\tan^2 x/2} \end{align*}
= \begin{align*}\int \frac{ \sec^2 {x/2}\; dx}{2 \tan^2 x/2 + 4 \tan {x/2} + 4} dx \end{align*}
=\frac{1}{2} \begin{align*}\int \frac{ \sec^2 {x/2} \;dx}{ \tan^2 x/2 + 2 \tan {x/2} + 2} dx \end{align*}
put $\tan x/2 = t \; \; \; \ ; \ \therefore \frac{1}{2} \sec^2 \; x/2 \;dx = dt$
\therefore I = \frac{2}{2} \begin{align*}\int \frac{dt} { t^2+ 2t+2 } \end{align*}
= \begin{align*}\int \frac{dt}{(t+1)^2 +1} \end{align*}
$= \tan^{-1} (1+t) + C$
$= \tan^{-1} (1+t) + C$
$= \tan^{-1} (1+ \tan^{-1} \frac{x}{2}) + C$
$\implies I = \tan^{-1} (1+ \tan^{-1} \frac{x}{2}) + C$