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?
 A: Note that
$$\int \cos(x)\log\left(\frac{9}{6-\sin(x)}\right) \, dx=\int\cos x\log9 \, dx-\int\cos x\log(6-\sin x) \, dx $$
The first integral is straight forward. The second can be approached by making the substitution $t=6-\sin x$ and then integrating by parts.
A: First part: $\int \cos(x)\,dx$, trivial. Second part: $\int \cos(x)\log(6-\sin(x))\,dx $ can be computed from:
$$ \int \log(6-u)\,du = u\log(6-u)+\int \frac{u}{6-u}\,du $$
(almost trivial), since $\cos = \sin'$.
A: Let $t = \sin x \implies I = \displaystyle \int (\ln 9 - \ln(6-t))dt= t\ln 9 - \displaystyle \int \ln(6-t)dt$. From this you can do integration by parts by letting $u = \ln(6-t), v = t$.
A: $$\int  \cos { \left( x \right)  } \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) dx=\int { \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) d\left( \sin { \left( x \right)  }  \right) = }  }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) -\frac { 1 }{ 9 } \int { \left( 6-\sin { \left( x \right)  }  \right) \sin { \left( x \right)  } dx } = }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) -\frac { 2 }{ 3 } \int { \sin { \left( x \right) dx+\frac { 1 }{ 9 } \int { \sin ^{ 2 }{ x } dx }  }  } = }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) -\frac { 2 }{ 3 } \int { \sin { \left( x \right) dx+\frac { 1 }{ 18 } \int { \left( 1-\cos { \left( 2x \right)  }  \right)  } dx }  } = }  } \\ =\sin { \left( x \right) \ln { \left( \frac { 9 }{ 6-\sin { \left( x \right)  }  }  \right) +\frac { 2 }{ 3 } \cos { \left( x \right)  } +\frac { 1 }{ 18 } \left( x-\frac { \sin { \left( 2x \right)  }  }{ 2 }  \right)  } +C } $$
