Mathematics

The value of integral $$\displaystyle \int _{ 0 }^{ \pi  }{ \frac { { x }^{ 2 }\sin { x }  }{ \left( 2x-\pi  \right) \left( 1+\cos ^{ 2 }{ x }  \right)  }  } dx$$ is equal to


ANSWER

$$\displaystyle\frac { { \pi }^{ 2 } }{ 4 } $$


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

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

$$\displaystyle\Rightarrow 2I=\pi \int _{ 0 }^{ \pi  }{ \frac { \sin { x }  }{ 1+\cos ^{ 2 }{ x }  }  } dx=-\pi \int _{ 1 }^{ -1 }{ \frac { dt }{ 1+{ t }^{ 2 } }  }$$   [putting $$\cos { x } =t\Rightarrow \sin { x } dx=-dt$$]

$$\displaystyle=\pi { \left[ \tan ^{ -1 }{ t }  \right]  }_{ -1 }^{ 1 }=\pi \left( \frac { \pi  }{ 4 } +\frac { \pi  }{ 4 }  \right) =\frac { { \pi  }^{ 2 } }{ 2 }$$
$$\displaystyle\therefore I=\frac { { \pi  }^{ 2 } }{ 4 } $$
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Single Correct Medium Published on 17th 09, 2020
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