Mathematics

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


ANSWER


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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