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Question: \(\int _ { 0 } ^ { \pi / 2 } \frac { \cos x } { ( 1 + \sin x ) ( 2 + \sin x ) } d x =\)...

∫0π/2cos⁡x(1+sin⁡x)(2+sin⁡x)dx=\int _ { 0 } ^ { \pi / 2 } \frac { \cos x } { ( 1 + \sin x ) ( 2 + \sin x ) } d x =

A

log⁡43\log \frac { 4 } { 3 }

B

log⁡13\log \frac { 1 } { 3 }

C

log⁡34\log \frac { 3 } { 4 }

D

None of these

Answer

log⁡43\log \frac { 4 } { 3 }

Explanation

Solution

Put sin⁡x=t⇒cos⁡xdx=dt\sin x = t \Rightarrow \cos x d x = d t so that reduced integral is ∫01(11+t−12+t)dt=[log⁡(1+t)−log⁡(2+t)]01\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 1 + t } - \frac { 1 } { 2 + t } \right) d t = [ \log ( 1 + t ) - \log ( 2 + t ) ] _ { 0 } ^ { 1 }

=log⁡23−log⁡12=log⁡43= \log \frac { 2 } { 3 } - \log \frac { 1 } { 2 } = \log \frac { 4 } { 3 }.