Elegant solution for $\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy$ I have the following integral: $\int {\frac{\cos(y)}{\sin^2(y)+\sin(y)-6}}dy$
I already know the solution, but it needs three substitutions. Is there a simpler, more elegant way to go about this?

$$\int { \frac { \cos { \left( x \right)  } dx }{ \sin ^{ 2 }{ \left( x \right)  } +\sin { \left( x \right)  } -6 }  } =\int { \frac { d\sin { \left( x \right)  }  }{ \left( \sin { \left( x \right) +3 }  \right) \left( \sin { \left( x \right) -2 }  
\right)  }  } $$
Let $t=\sin { \left( x \right)  } $ then
$$\frac { 1 }{ \left( t+3 \right) \left( t-2 \right)  } =\frac { A }{ t+3 } +\frac { B }{ t-2 } $$
$$1=A\left( t-2 \right) +B\left( t+3 \right) \\ 1=\left( A+B \right) t-2A+3B\\ \begin{cases} A+B=0 \\ -2A+3B=1 \end{cases}\Rightarrow \begin{cases} A=-B \\ 2B+3B=1 \end{cases}\Rightarrow B=\frac { 1 }{ 5 } ,A=-\frac { 1 }{ 5 } $$ so 
$\frac { 1 }{ \left( t+3 \right) \left( t-2 \right)  } =\frac { 1 }{ 5 } \left( \frac { 1 }{ t-2 } -\frac { 1 }{ t+3 }  \right) $
$$\\ \\ \\ \frac { 1 }{ 5 } \int { \left( \frac { 1 }{ t-2 } -\frac { 1 }{ t+3 }  \right) dt=\frac { 1 }{ 5 } \int { \frac { dt }{ t-2 }  } -\frac { 1 }{ 5 } \int { \frac { dt }{ t+3 } = }  } $$ $$=\frac { 1 }{ 5 } \int { \frac { d\left( t-2 \right)  }{ t-2 }  } -\frac { 1 }{ 5 } \int { \frac { d\left( t+3 \right)  }{ t+3 } =\frac { 1 }{ 5 } \left[ \ln { \left| t-2 \right|  } -\ln { \left| t+3 \right|  }  \right] =\frac { 1 }{ 5 } \ln { \left| \frac { t-2 }{ t+3 }  \right| +C }  }  $$
Hence $t=\sin { \left( x \right)  } $,$$\int { \frac { \cos { \left( x \right)  } dx }{ \sin ^{ 2 }{ \left( x \right)  } +\sin { \left( x \right)  } -6 }  } =\frac { 1 }{ 5 } \ln { \left| \frac { \sin { \left( x \right) -2 }  }{ \sin { \left( x \right) +3 }  }  \right|  } +C$$
