Mathematics

If $$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \sin { t }  }{ 1+t } dt } =\alpha $$, then the value of the integral $$\displaystyle \int _{ 4\pi -2 }^{ 4\pi  }{ \frac { \sin { t/2 }  }{ 4\pi +2-t } dt } $$ in terms of $$\alpha$$ is given by


ANSWER

$$-\alpha$$


SOLUTION
$$\displaystyle \int _{ 4\pi -2 }^{ 4\pi  }{ \frac { \sin { t/2 }  }{ 4\pi +2-t } dt } =\frac { 1 }{ 2 } \int _{ 4\pi -2 }^{ 4\pi  }{ \frac { \sin { t/2 }  }{ 1+\left( 2\pi -\dfrac { t }{ 2 }  \right)  } dt } $$

$$\displaystyle =2.\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { \sin { \left( 2\pi -u \right)  }  }{ 1+u } du } $$

$$($$ Substitute $$\displaystyle 2\pi -\frac { t }{ 2 } =u\Rightarrow dt=-2du)$$

$$\displaystyle =\int _{ 0 }^{ 1 }{ \frac { \sin { u }  }{ 1+u } du } =-\int _{ 0 }^{ 1 }{ \frac { \sin { t }  }{ 1+t } dt } =-\alpha $$
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Single Correct Hard Published on 17th 09, 2020
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