Mathematics

Solve:$$\displaystyle \int \dfrac{sin x. \cos x}{a\cos^2x+b\sin^2 x}.dx$$


SOLUTION
Given the integral,
$$\int { \dfrac { \cos { x } \sin { x }  }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x }  }  } dx$$
Let us assume,
$$u=b\sin ^{ 2 }{ x } +a\cos ^{ 2 }{ x } \\ \Rightarrow \dfrac { du }{ dx } =2b\cos { x } \sin { x } -2a\cos { x } \sin { x } \\ \Rightarrow \dfrac { du }{ dx } =\cos { x } \sin { x } (2b-2a)\\ \Rightarrow \cos { x } \sin { x } dx=\dfrac { 1 }{ 2b-2a } du$$
Substituting these values in the given integral we get,
$$\int { \dfrac { \cos { x } \sin { x }  }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x }  }  } dx\\ =\int { \dfrac { 1 }{ (2b-2a)u } du } \\ =\dfrac { 1 }{ 2b-2a } \int { \dfrac { 1 }{ u } du } \\ =\dfrac { 1 }{ 2b-2a } \ln { (u) } \\ =\dfrac { \ln { (u) }  }{ 2b-2a } \\ =\dfrac { \ln { (a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x } ) }  }{ 2b-2a } \\ \therefore \int { \dfrac { \cos { x } \sin { x }  }{ a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x }  }  } dx=\dfrac { \ln { \left( \left| a\cos ^{ 2 }{ x } +b\sin ^{ 2 }{ x }  \right|  \right)  }  }{ 2b-2a } +C.$$
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Subjective Medium Published on 17th 09, 2020
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