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Question: \[\int_{}^{}{\frac{\cos x}{(1 + \sin x)(2 + \sin x)}dx =}\]...

∫cos⁡x(1+sin⁡x)(2+sin⁡x)dx=\int_{}^{}{\frac{\cos x}{(1 + \sin x)(2 + \sin x)}dx =}

A

log⁡[(1+sin⁡x)(2+sin⁡x)]+c\log\lbrack(1 + \sin x)(2 + \sin x)\rbrack + c

B

log⁡[2+sin⁡x1+sin⁡x]+c\log\left\lbrack \frac{2 + \sin x}{1 + \sin x} \right\rbrack + c

C

log⁡[1+sin⁡x2+sin⁡x]+c\log\left\lbrack \frac{1 + \sin x}{2 + \sin x} \right\rbrack + c

D

None of these

Answer

log⁡[1+sin⁡x2+sin⁡x]+c\log\left\lbrack \frac{1 + \sin x}{2 + \sin x} \right\rbrack + c

Explanation

Solution

Put sin⁡x=t\sin x = t ⇒cos⁡xdx=dt\Rightarrow \cos xdx = dt

∫cos⁡xdx(1+sin⁡x)(2+sin⁡x)=∫dt(1+t)(2+t)=∫[1t+1−1t+2]dt\int \frac { \cos x d x } { ( 1 + \sin x ) ( 2 + \sin x ) } = \int \frac { d t } { ( 1 + t ) ( 2 + t ) } = \int \left[ \frac { 1 } { t + 1 } - \frac { 1 } { t + 2 } \right] d t =log⁡(t+1)−log⁡(t+2)+c= \log(t + 1) - \log(t + 2) + c =log⁡[t+1t+2]+c\log\left\lbrack \frac{t + 1}{t + 2} \right\rbrack + c =log⁡[1+sin⁡x2+sin⁡x]+c= \log \left[ \frac { 1 + \sin x } { 2 + \sin x } \right] + c