Mathematics

The value of the definite integral
$$\overset { { a }_{ 1 } }{ \underset { { a }_{ 2 } }{ \int { \frac { d\theta  }{ 1+tan\theta  }  }  }  } =\frac { 501\pi  }{ K } $$ where $$\ a _{ 2 }=\quad \frac { 1003\pi  }{ 2008 } $$ and $${ \ a  }_{ 1 }=\frac { \pi  }{ 2008 } $$ The value of K equalls


ANSWER

2008


SOLUTION
$$\begin{array}{l} \int  _{ { a_{ 2 } } }^{ { a_{ 1 } } }{ \frac { { d\theta  } }{ { 1+\tan  \theta  } }  } \\ I=\int  _{ \frac { \pi  }{ { 2008 } }  }^{ \frac { { 1003\pi  } }{ { 2008 } }  }{ \frac { { d\theta  } }{ { 1+\tan  \theta  } }  } \\ I=\int  _{ \frac { \pi  }{ { 2008 } }  }^{ \frac { { 1003\pi  } }{ { 2008 } }  }{ \frac { { \tan  \theta  } }{ { 1+\tan  \theta  } }  }d\theta  \\ 2I=\int  _{ \frac { \pi  }{ { 2008 } }  }^{ \frac { { 1003\pi  } }{ { 2008 } }  }{ d\theta  } \\ 2I=\frac { { 1002\pi  } }{ { 2008 } }  \\ I=\frac { { 501 } }{ { 2008 } } \pi  \\ k=2008 \end{array}$$
Option $$D$$ is correct answer.
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