Mathematics

The value of $$\int_{-\dfrac{\pi}{2}} ^ {\dfrac{\pi}{2}} \dfrac{x^2\cos x}{1+e^x}dx$$ is equal to 


SOLUTION
Let $$I=\int _{ \cfrac { -\pi  }{ 2 }  }^{ \cfrac { \pi  }{ 2 }  }{ \cfrac { { x }^{ 2 }\cos { x }  }{ 1+{ e }^{ x } } dx } $$
Using $$\int _{ -a }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ (f(x)+f(-x))dx } $$ , we get 
$$I=\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ \cfrac { { x }^{ 2 }\cos { x } (1+{ e }^{ x })dx }{ (1+{ e }^{ x }) }  } =-\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ { x }^{ 2 }\cos { x } dx } $$
using product rule, we get
$$I={ x }^{ 2 }\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ \cos { x } dx } -\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ [(\cfrac { d{ x }^{ 2 } }{ dx } )\int { \cos { x } dx } ]dx } \\ I={ x }^{ 2 }{ [\sin { x } ] }_{ 0 }^{ \cfrac { \pi  }{ 2 }  }-\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ 2x\sin { x } dx } \\ I=\cfrac { { \pi  }^{ 2 } }{ 4 } -2[x\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ \sin { x } dx } -\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ [\cfrac { dx }{ dx } \int { \sin { x } dx } ]dx } ]\\ I=\cfrac { { \pi  }^{ 2 } }{ 4 } -2[0+\int _{ 0 }^{ \cfrac { \pi  }{ 2 }  }{ \cos { x } dx } ]\\ I=\cfrac { { \pi  }^{ 2 } }{ 4 } -2(1)=\cfrac { { \pi  }^{ 2 } }{ 4 } -2$$
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Subjective Medium Published on 17th 09, 2020
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