Mathematics

Solve: $$\int\dfrac{1}{(\sin x-2 \cos x)(2\sin x+\cos x)}dx$$


SOLUTION
Let, $$I=\int { \dfrac { 1 }{ (\sin { x } -2\cos { x) } (2\sin { x } +\cos { x } ) }  } dx\\ =\int { \dfrac { 1 }{ 2{ \sin }^ { 2 }x-3\sin { x\cos { x }  } -2{ \cos } ^{ 2 }x }  } dx\\ =\int { \dfrac { 1 }{ -2({ \cos }^ { 2 }x-{ \sin }^ { 2 }x)-3\sin { x\cos { x }  }  }  } dx\\ =-\int { \dfrac { 1 }{ 2\cos { 2x } +\dfrac { 3 }{ 2 } \sin { 2x }  }  } dx$$
Now, let $$\tan { x } =t\\ \therefore dx=\dfrac { 2 }{ 1+{ t }^{ 2 } } dt,\cos { 2x } =\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ,\sin { 2x } =\dfrac { 2t }{ 1+{ t }^{ 2 } } $$
We can write, 
$$I=-\int { \dfrac { 1 }{ 2(\dfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } )+\dfrac { 3 }{ 2 } (\dfrac { 2t }{ 1+{ t }^{ 2 } } ) }  } \cdot (\dfrac { 2 }{ 1+{ t }^{ 2 } } dt)\\ =-\int { \dfrac { 1 }{ \dfrac { 2-2{ t }^{ 2 }+3t }{ 1+{ t }^{ 2 } }  }  } \cdot \dfrac { 2 }{ 1+{ t }^{ 2 } } dt\\ =\int { \dfrac { 2 }{ 2{ t }^{ 2 }-3t-2 }  } dt\\ =\int { \dfrac { 1 }{ { t }^{ 2 }-\dfrac { 3 }{ 2 } t-1 }  } dt\\ =\int { \dfrac { 1 }{ { t }^{ 2 }-\dfrac { 3 }{ 2 } t-1+\dfrac { 9 }{ 16 } -\dfrac { 9 }{ 16 }  }  } dt\\ =\int { \dfrac { 1 }{ ({ t }^{ 2 }-\dfrac { 3 }{ 2 } t+\dfrac { 9 }{ 16 } )-(\dfrac { 9 }{ 16 } +\dfrac { 16 }{ 16 } ) }  } dt\quad \quad \left[ putting,1=\dfrac { 16 }{ 16 }  \right] \\ =\int { \dfrac { 1 }{ ({ t-\dfrac { 3 }{ 2 }  })^{ 2 }-{ (\dfrac { 5 }{ 4 } ) }^{ 2 } }  } dt$$
We know that,
$$\int { \dfrac { 1 }{ ({ { x }^{ 2 }-{ a }^{ 2 }) } }  } dx=\dfrac { 1 }{ 2a } \log { \left[ \dfrac { x-a }{ x+a }  \right]  } +C$$
$$\therefore I=\dfrac { 1 }{ 2(\dfrac { 5 }{ 4 } ) } \log { \left[ \dfrac { (t-\dfrac { 3 }{ 2 } )-\dfrac { 5 }{ 4 }  }{ (t-\dfrac { 3 }{ 2 } )+\dfrac { 5 }{ 4 }  }  \right]  } +C\\ =\dfrac { 2 }{ 5 } \log { \left[ \dfrac { 4t-11 }{ 4t-1 }  \right]  } +C$$
Putting $$t=\tan { x } ,\\ \Rightarrow I=\dfrac { 2 }{ 5 } \log { \left[ \dfrac { 4\tan { x } -11 }{ 4\tan { x } -1 }  \right]  } +C\\ \therefore \int { \dfrac { 1 }{ (\sin { x } -2\cos { x) } (2\sin { x } +\cos { x } ) }  } dx=\dfrac { 2 }{ 5 } \log { \left[ \dfrac { 4\tan { x } -11 }{ 4\tan { x } -1 }  \right]  } +C.$$
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Subjective Medium Published on 17th 09, 2020
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