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Calculus Level 5

0πk=0sin(x(2k+1))2k+1dx=πab,\int _{ 0 }^{ \pi }{ \sum _{ k=0 }^{ \infty }{ \frac { \sin { \left( x\left( 2k+1 \right) \right) } }{ 2k+1 } } } dx= \frac{ \pi ^a } { b },

where a a and bb are integers. Find 10a+b10a+b.


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