If ∫(3+2cos⁡x)(2+3cos⁡x)2dx=P(x)A+BQ(x)+C then

∫3+2cos⁡x(2+3cos⁡x)2dx=∫(3+2cos⁡x)cosec2⁡x(2+3cos⁡x)2cosec2⁡xdx=∫3cosec2⁡x+2cotx ⁡cosec⁡x(2cosec⁡x+3cot⁡x)2 put 2cosec⁡x+3cot⁡x=t⇒-3cosec2⁡x+2cotx ⁡cosec⁡xdx=dt=∫−dtt2=1t+c=12cosec⁡x+3cot⁡x+c=sin⁡x2+3cos⁡x+cP(x)=sin⁡x, Q(x)=cos⁡x, A=2, B=3

If ∫(3+2cos⁡x)(2+3cos⁡x)2dx=P(x)A+BQ(x)+C then

∫3+2cos⁡x(2+3cos⁡x)2dx=∫(3+2cos⁡x)cosec2⁡x(2+3cos⁡x)2cosec2⁡xdx=∫3cosec2⁡x+2cotx ⁡cosec⁡x(2cosec⁡x+3cot⁡x)2 put 2cosec⁡x+3cot⁡x=t⇒-3cosec2⁡x+2cotx ⁡cosec⁡xdx=dt=∫−dtt2=1t+c=12cosec⁡x+3cot⁡x+c=sin⁡x2+3cos⁡x+cP(x)=sin⁡x, Q(x)=cos⁡x, A=2, B=3

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$If \int \frac{(3 + 2 \cos x)}{(2 + 3 \cos x)^{2}} d x = \frac{P (x)}{A + B Q (x)} + C then$

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$P^{2} (x) + Q^{2} (x) = 1$

$P^{2} (x) - Q^{2} (x) = 1$

AB = 6

AB = 1

answer is A, C.

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Detailed Solution

$\begin{array}{l} \int \frac{3 + 2 \cos x}{(2 + 3 \cos x)^{2}} dx \\ = \int \frac{(3 + 2 \cos x) {cosec}^{2} x}{(2 + 3 \cos x)^{2} {cosec}^{2} x} dx \\ = \int \frac{3 {cosec}^{2} x + 2 \cot x_{} cosec x}{(2 cosec x + 3 \cot x)^{2}} \end{array} p u t 2 cosec x + 3 \cot x = t \Rightarrow - (3 {cosec}^{2} x + 2 {cotx}_{} cosec x) dx = dt$

$\begin{array}{l} = \int \frac{- dt}{t^{2}} \\ = \frac{1}{t} + c \\ = \frac{1}{2 cosec x + 3 \cot x} + c \\ = \frac{\sin x}{2 + 3 \cos x} + c \end{array}$