Mathematics

$$\int { \dfrac { \sin { 2x }  }{ \left( 1+\sin { x }  \right) \left( 2+\sin { x }  \right)  }  } dx$$


SOLUTION

Let $$I=\int { \dfrac { sin2x }{ \left( 1+sinx \right) \left( 2+sinx \right)  }  } dx$$

 

$$I=\int { \dfrac { sin2xcosx }{ \left( 1+sinx \right) \left( 2+sinx \right)  }  } dx$$

 

put $$sinx=t$$

 

$$cosxdx=dt$$

 

$$I=\int { \dfrac { 2tdt }{ \left( 1+t \right) \left( 2+t \right)  }  } $$

 

$$\int { \dfrac { 2tdt }{ \left( 1+t \right) \left( 2+t \right)  }  } =\dfrac { A }{ \left( 1+t \right)  } +\dfrac { B }{ \left( 2+t \right)  } $$

 

$$2t=A\left( 2+t \right) +B\left( 1+t \right) $$

 

$$2t=2A+At+B+Bt$$

 

$$2t=\left( A+B \right) t+2A+B$$

 

comparing the coefficients,

 

$$A+B=2\Longrightarrow (1)$$

 

comparing constants,

 

$$2A+B=0\Longrightarrow (2)$$

 

Subtracting (1) from (2)

 

$$A=-2$$

 

$$B=4$$

 

$$\dfrac { 2t }{ \left( 1+t \right) \left( 2+t \right)  } =\dfrac { -2 }{ \left( 1+t \right)  } +\dfrac { 4 }{ \left( 2+t \right)  } $$

 

By integrating,

 

$$\int { \dfrac { 2t }{ \left( 1+t \right) \left( 2+t \right)  }  } dt=-2\int { \dfrac { dt }{ \left( 1+t \right)  } +4\int { \dfrac { dt }{ \left( 2+t \right)  }  }  } =-2log\left| 1+t \right| +4log\left| 2+t \right| +C$$

 

$$I=-2log\left| 1+sinx \right| +4log\left| 2+t \right| +C$$

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