Mathematics

# Solve:$\int \dfrac{1}{5-4 \cos x}.dx$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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