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I(n)=∫0π2sin2nx dxX=∑r=1∞I(r)(2rr)I(n)= \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin ^{ 2n }{ x } } \ dx \qquad \qquad X=\sum _{ r=1 }^{ \infty }{ \frac { I(r) }{ { \left( \begin{matrix} 2r \\r \end{matrix} \right) } } } I(n)=∫02πsin2nx dxX=r=1∑∞(2rr)I(r)
Given the above, find ⌊104X⌋\left\lfloor { 10 }^{ 4 }X \right\rfloor ⌊104X⌋
Notation:
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