Mathematics

Evaluate:
$$\displaystyle \int \dfrac{1}{5-4 \cos x}.dx$$


SOLUTION
$$\quad \int { \dfrac { 1 }{ 5-4\cos x } dx\quad  } \\ =\int { \dfrac { dx }{ 5\left( { \cos }^{ 2 }\dfrac { x }{ 2 } +{ \sin }^{ 2 }\dfrac { x }{ 2 }  \right) +4\left( { \cos }^{ 2 }\dfrac { x }{ 2 } -{ \sin }^{ 2 }\dfrac { x }{ 2 }  \right)  }  } \\ =\int { \dfrac { dx }{ 5{ \cos }^{ 2 }\dfrac { x }{ 2 } +5{ \sin }^{ 2 }\dfrac { x }{ 2 } +4{ \cos }^{ 2 }\dfrac { x }{ 2 } -4{ \sin }^{ 2 }\dfrac { x }{ 2 }  }  } \\ =\int { \dfrac { dx }{ 9{ \cos }^{ 2 }\dfrac { x }{ 2 } +{ \sin }^{ 2 }\dfrac { x }{ 2 }  }  } \\ =\int { \dfrac { { \sec }^{ 2 }\dfrac { x }{ 2 }  }{ 9+{ \tan }^{ 2 }\dfrac { x }{ 2 }  } dx\quad  } \\ Now,\quad let\\ \tan\dfrac { x }{ 2 } =t\\ differentiating\quad w.r.t\quad x,\quad we\quad get\\ \quad \quad { \sec }^{ 2 }\dfrac { x }{ 2 } \times \dfrac { 1 }{ 2 } dx=dt\\ \Rightarrow { \sec }^{ 2 }\dfrac { x }{ 2 } dx=2dt\\ now\quad putting\quad these\quad values\quad in\quad the\quad above\quad equation\quad we\quad get,\\ \quad \quad \int { \dfrac { { \sec }^{ 2 }\dfrac { x }{ 2 }  }{ 9+{ \tan }^{ 2 }\dfrac { x }{ 2 }  } dx\quad  } \\ =2\int { \dfrac { dt }{ 9+{ t }^{ 2 } }  } \\ =2\int { \dfrac { dt }{ { \left( 3 \right)  }^{ 2 }+{ \left( t \right)  }^{ 2 } }  } \\ =\dfrac { 2 }{ 3 } { \tan }^{ -1 }\left( \dfrac { t }{ 3 }  \right) \\ =\dfrac { 2 }{ 3 } { \tan }^{ -1 }\left( \dfrac { \tan\dfrac { x }{ 2 }  }{ 3 }  \right) +C$$
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Subjective Medium Published on 17th 09, 2020
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