Mathematics

The value of the integral $$\displaystyle \int_0^{2 \pi} \frac{sin  2 \theta}{a - b  cos  \theta} d \theta$$ when a > b > 0, is


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SOLUTION
Let $$I=\int _{ 0 }^{ 2\pi  }{ \cfrac { sin2\theta  }{ a-bcos\theta  } d\theta  } $$   ...(1)
Using property $$\int _{ 0 }^{ a }{ f\left( x \right) dx } =\int _{ 0 }^{ a }{ f\left( a-x \right) dx } $$
$$I=\int _{ 0 }^{ 2\pi  }{ \cfrac { sin2\left( 2\pi -\theta  \right)  }{ a-bcos\left( 2\pi -\theta  \right)  } d\theta  } =\int _{ 0 }^{ 2\pi  }{ \cfrac { sin\left( 4\pi -2\theta  \right)  }{ a-bcos\theta  } d\theta  } $$
$$=-\int _{ 0 }^{ 2\pi  }{ \cfrac { sin2\theta  }{ a-bcos\theta  } d\theta  } $$   ...(2)
Adding (1) and (2)
$$2I=0\Rightarrow I=0$$

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Single Correct Medium Published on 17th 09, 2020
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