<s>[INST] Solving $\int\cos(x)\ln\left(\frac{9}{6-\sin(x)}\right)\,dx $ I have:
$$\int\cos(x)\ln\left(\frac{9}{6-\sin(x)}\right)\,dx $$
I've tried by parts but without results, i don't know how to start, any tips?
 [/INST] $$\int  \cos { \left( x \right)  } \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) dx=\int { \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) d\left( \sin { \left( x \right)  }  \right) = }  }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) -\frac { 1 }{ 9 } \int { \left( 6-\sin { \left( x \right)  }  \right) \sin { \left( x \right)  } dx } = }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) -\frac { 2 }{ 3 } \int { \sin { \left( x \right) dx+\frac { 1 }{ 9 } \int { \sin ^{ 2 }{ x } dx }  }  } = }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) -\frac { 2 }{ 3 } \int { \sin { \left( x \right) dx+\frac { 1 }{ 18 } \int { \left( 1-\cos { \left( 2x \right)  }  \right)  } dx }  } = }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) +\frac { 2 }{ 3 } \cos { \left( x \right)  } +\frac { 1 }{ 18 } \left( x-\frac { \sin { \left( 2x \right)  }  }{ 2 }  \right)  } +C } $$
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