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intergation of 1/(a+bcosX) please step by steps there are lots of solution in net but i want detail solution

intergation of 1/(a+bcosX) please step by steps
there are lots of solution in net but i want detail solution

Grade:12th pass

1 Answers

Ajay
209 Points
7 years ago
Detailed solution below-------------------------------------------------------------------------------------
Let\quad tan\frac { x }{ 2 } \quad =\quad t\\ \frac { 1 }{ 2 } { sec }^{ 2 }\frac { x }{ 2 } dx\quad =\quad dt\quad =>\quad \frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } { tan }^{ 2 }\frac { x }{ 2 } +1 \right) dx\quad =\quad dt\\ or\quad dx=\quad \frac { 2dt }{ { t }^{ 2 }+1 } -------(1)\\ Also\quad cosx\quad =\quad 2{ cos }^{ 2 }\frac { x }{ 2 } -1\quad =\quad \frac { 2 }{ sec^{ 2 }\frac { x }{ 2 } } -1\\ cosx\quad =\quad \frac { 2 }{ 1+tan^{ 2 }\frac { x }{ 2 } } -1\quad =\quad \frac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ----(2)\\ From\quad (1)\quad \quad and\quad (2)\quad the\quad integral\quad becomes\\ \int { \frac { 2dt }{ 1+{ t }^{ 2 } } \left( \frac { 1 }{ a+b\left( \frac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } \right) } \right) }=\quad \int { \frac { 2dt }{ (a-b){ t }^{ 2 }+(a+b) } } \\ =\frac { 2 }{ \left( a-b \right) } \int { \frac { 2dt }{ { t }^{ 2 }+\frac { (a+b) }{ \left( a-b \right) } } } \\ =\quad \frac { 2 }{ \left( a-b \right) } \frac { \sqrt { a-b } }{ \sqrt { a+b } } { tan }^{ -1 }\left( t\frac { \sqrt { a-b } }{ \sqrt { a+b } } \right) \quad +\quad C\\ =\quad \frac { 2 }{ \sqrt { { a }^{ 2 }-b^{ 2 } } } { tan }^{ -1 }\left( tan\frac { x }{ 2 } \frac { \sqrt { a-b } }{ \sqrt { a+b } } \right) \quad +\quad C

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