Question

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Answer 1

Answer:

$\dfrac{4}{ \sqrt{3} } \: {tan}^{ - 1} ( \dfrac{tan \dfrac{x}{2} }{ \sqrt{3} } ) + log(2 + cosx) + c$

Step-by-step explanation:

$\bold{Solution}\longrightarrow \\ I = \displaystyle\int \dfrac{2 - sinx}{2 + cosx} \: dx \\ = \displaystyle\int\dfrac{2}{2 + cosx} \: dx - \displaystyle\int \dfrac{sinx}{2 + cosx} dx \\ I = I(1) - I(2)..........(1) \\ now \\ I(1) = \displaystyle\int\dfrac{2}{2 + cosx} dx \\ = \displaystyle\int \dfrac{2}{2 + \dfrac{1 - {tan}^{2} \dfrac{x}{2} }{1 + {tan}^{2} \dfrac{x}{2} } } \: dx \\ = \displaystyle\int \dfrac{2}{ \dfrac{2 + 2 {tan}^{2} \dfrac{x}{2} + 1 - {tan}^{2} \dfrac{x}{2} }{1 + {tan}^{2} \dfrac{x}{2} } } \: dx \\ = \displaystyle\int\dfrac{2(1 + {tan}^{2} \dfrac{x}{2} ) }{3 + {tan}^{2} \dfrac{x}{2} } \: dx \\ =\displaystyle\int \dfrac{2 {sec}^{2} \dfrac{x}{2} }{3 + {tan}^{2} \dfrac{x}{2} } \: dx \\ now \\ let \: \: tan \dfrac{x}{2} = t \\ differentiating \: with \: repect \: to \: x \\ {sec}^{2} \dfrac{x}{2} ( \dfrac{1}{2} ) \: dx = dt \\ {sec}^{2} \dfrac{x}{2} \: dx = 2 \: dt \\ I(1) = \displaystyle\int \dfrac{2}{3 + {t}^{2} } \: \: 2dt \\ = \displaystyle\int \dfrac{4 \: dt}{ {( \sqrt{3}) }^{2} + {t}^{2} } \\ = 4 \dfrac{1}{ \sqrt{3} } {tan}^{ - 1} ( \dfrac{t}{ \sqrt{3} } ) + c \\ I(1) = \dfrac{4}{ \sqrt{3} } {tan}^{ - 1} ( \dfrac{tan \dfrac{x}{2} }{ \sqrt{3} } )$

$now$

$I(2) = \displaystyle\int \dfrac{sinx}{2 + cosx} dx \\ let \\ 2 + cosx = u \\ differentiating \: both \: sides \\ (0 - sinx)dx = du \\ sinx \: dx = - du \\ I(2) = \displaystyle\int \dfrac{ - du}{u} \\ = - logu \\ = - log(2 + cosx)$

$now \\ I = I(1) - I(2) \\ I = \dfrac{4}{ \sqrt{3} } {tan}^{ - 1} ( \dfrac{tan \dfrac{x}{2} }{ \sqrt{3} } ) + log(2 + cosx) + c$