Mathematics

# Evaluate: $\displaystyle \int _{ 0 }^{\frac{\pi}{2} }{ \dfrac { \sin { x } -\cos { x } }{ 1+\sin { x } \cos { x } } } dx$

##### SOLUTION
$\displaystyle{I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } -\cos x }{ 1+\sin x\cos x } } dx.........(i)\\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { \left( \frac { \pi }{ 2 } -x \right) } -\cos \left( \frac { \pi }{ 2 } -x \right) }{ 1+\sin \left( \frac { \pi }{ 2 } -x \right) \cos \left( \frac { \pi }{ 2 } -x \right) } } dx\\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \cos x-\sin x }{ 1+\sin x\cos x } } dx\\ I=-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sin { x } -\cos x }{ 1+\sin x\cos x } } dx.........(ii)}$

Adding $(i)$ and $(ii)$

$2I=0$

$I=0$

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One Word Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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