Mathematics

Evaluate: $$\int\limits_\pi ^{\tfrac{{5\pi }}{4}} {\dfrac{{\sin2x.dx}}{{\cos {^4}x + \sin {^4}x}}  } $$


ANSWER

$$\dfrac{\pi }{4}$$


SOLUTION
$$\int _{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }{ \dfrac { \sin { 2x } dx }{ \cos ^{ 4 }{ x } +\sin ^{ 4 }{ x }  }  } =\int _{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }{ \dfrac { \sin { 2x } dx }{ { \left( \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x }  \right)  }^{ 2 }-2\sin ^{ 2 }{ x } \cos ^{ 2 }{ x }  }  } $$

$$=\int _{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }{ \cfrac { \sin { 2x } dx }{ { 1-\dfrac { 1 }{ 2 } \left( \sin ^{ 2 }{ 2x }  \right)  } }  } =\int _{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }{ \cfrac { \sin { 2x } dx }{ 1+1-\sin ^{ 2 }{ 2x }  }  } =\int _{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }{ \cfrac { \sin { 2x }  }{ 1+\cos ^{ 2 }{ 2x }  }  } dx$$

$$\cos { 2x } =t\Rightarrow -2\sin { 2x } dx=dt$$

$$\int _{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }{ \dfrac { -1 }{ 1+{ t }^{ 2 } }  } dt={ \left[ -\tan ^{ -1 }{ t }  \right]  }_{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }={ \left[ -\tan ^{ -1 }{ \cos { 2x }  }  \right]  }_{ \pi  }^{ \dfrac { 5\pi  }{ 4 }  }=0+\dfrac { \pi  }{ 4 } =\dfrac { \pi  }{ 4 } $$
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Single Correct Medium Published on 17th 09, 2020
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