Question

$\int _ { 0 } ^ { \pi / 2 } \frac { \sin x \cos x d x } { \cos ^ { 2 } x + 3 \cos x + 2 } =$

• A. $\log \left( \frac { 8 } { 9 } \right)$
• B. $\log \left( \frac { 9 } { 8 } \right)$
• C. $\log ( 8 \times 9 )$
• D. None of these

Answer 1

Answer: (B)

$\log \left( \frac { 9 } { 8 } \right)$

Answer 2

Let $I = \int _ { 0 } ^ { \pi / 2 } \frac { \sin x \cos x \cdot d x } { \cos ^ { 2 } x + 3 \cos x + 2 }$

We put $\cos x = t \Rightarrow - \sin x d x = d t$ then

$= [ 2 \log ( t + 2 ) - \log ( t + 1 ) ] _ { 0 } ^ { 1 }$ $= [ 2 \log 3 - \log 2 - 2 \log 2 ]$

$= [ 2 \log 3 - 3 \log 2 ] = [ \log 9 - \log 8 ] = \log \left( \frac { 9 } { 8 } \right)$ .