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

∫0π/2cos⁡x−sin⁡x1+sin⁡xcos⁡xdx=\int _ { 0 } ^ { \pi / 2 } \frac { \cos x - \sin x } { 1 + \sin x \cos x } d x =

A

2

B

C

0

D

None of these

Answer

0

Explanation

Solution

∫0π/2cos⁡x−sin⁡x1+sin⁡xcos⁡xdx=I\int _ { 0 } ^ { \pi / 2 } \frac { \cos x - \sin x } { 1 + \sin x \cos x } d x = I …..(i)

Now I=∫0π/2cos⁡(π2−x)−sin⁡(π2−x)1+sin⁡(π2−x)cos⁡(π2−x)dxI = \int _ { 0 } ^ { \pi / 2 } \frac { \cos \left( \frac { \pi } { 2 } - x \right) - \sin \left( \frac { \pi } { 2 } - x \right) } { 1 + \sin \left( \frac { \pi } { 2 } - x \right) \cos \left( \frac { \pi } { 2 } - x \right) } d x

= ∫0π/2sin⁡x−cos⁡x1+sin⁡xcos⁡xdx\int _ { 0 } ^ { \pi / 2 } \frac { \sin x - \cos x } { 1 + \sin x \cos x } d x .....(ii)

On adding, 2I=0⇒I=02 I = 0 \Rightarrow I = 0.