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Question: \(\int \frac { 3 + 2 \cos x } { ( 2 + 3 \cos x ) ^ { 2 } } d x\) is equal to...

∫3+2cos⁡x(2+3cos⁡x)2dx\int \frac { 3 + 2 \cos x } { ( 2 + 3 \cos x ) ^ { 2 } } d x is equal to

A

(sin⁡x3cos⁡x+2)+c\left( \frac { \sin x } { 3 \cos x + 2 } \right) + c

B
C

(2cos⁡x3cos⁡x+2)+c\left( \frac { 2 \cos x } { 3 \cos x + 2 } \right) + c

D

(2sin⁡x3sin⁡x+2)+c\left( \frac { 2 \sin x } { 3 \sin x + 2 } \right) + c

Answer

(sin⁡x3cos⁡x+2)+c\left( \frac { \sin x } { 3 \cos x + 2 } \right) + c

Explanation

Solution

Let I = ∫3+2cos⁡x(2+3cos⁡x)2dx\int \frac { 3 + 2 \cos x } { ( 2 + 3 \cos x ) ^ { 2 } } d x Multiplying Nr. & Dr. by cosec2x ⇒ I =∫(3cosec⁡2x+2cot⁡xcosec⁡x)(2cosec⁡x+3cot⁡x)2\int \frac { \left( 3 \operatorname { cosec } ^ { 2 } x + 2 \cot x \operatorname { cosec } x \right) } { ( 2 \operatorname { cosec } x + 3 \cot x ) ^ { 2 } }dx

= –∫−3cosec⁡2x−2cot⁡xcosec⁡x(2cosec⁡x+3cot⁡x)2\int \frac { - 3 \operatorname { cosec } ^ { 2 } x - 2 \cot x \operatorname { cosec } x } { ( 2 \operatorname { cosec } x + 3 \cot x ) ^ { 2 } }dx

= 12cosec⁡x+3cot⁡x\frac { 1 } { 2 \operatorname { cosec } x + 3 \cot x } = (sin⁡x2+3cos⁡x)+c\left( \frac { \sin x } { 2 + 3 \cos x } \right) + c

Alternative Solution:

I = ∫3sin⁡2x+3cos⁡2x+2cos⁡x(2+3cos⁡x)2dx\int \frac { 3 \sin ^ { 2 } x + 3 \cos ^ { 2 } x + 2 \cos x } { ( 2 + 3 \cos x ) ^ { 2 } } d x

= ∫cos⁡x(2+3cos⁡x)dx+∫3sin⁡x⋅sin⁡x(2+3cos⁡x)2dx\int \frac { \cos x } { ( 2 + 3 \cos x ) } d x + \int \frac { 3 \sin x \cdot \sin x } { ( 2 + 3 \cos x ) ^ { 2 } } d x

= ∫cos⁡x2+3cos⁡xdx+sin⁡x2+3cos⁡x−∫cos⁡x2+3cos⁡xdx=sin⁡x2+3cos⁡x+c\int \frac { \cos x } { 2 + 3 \cos x } d x + \frac { \sin x } { 2 + 3 \cos x } - \int \frac { \cos x } { 2 + 3 \cos x } d x = \frac { \sin x } { 2 + 3 \cos x } + c