Mathematics

# Solve:-$\displaystyle\int {\dfrac{1}{{\cos \lambda + \cos x}}} dx$

##### SOLUTION
$\cos x = \dfrac{{1 - {{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}}$

$\displaystyle\int {\dfrac{{1 + {{\tan }^2}x/2}}{{\cos x\left( {1 + {{\tan }^2}x/2} \right) + 1 - {{\tan }^2}x/2}}dx}$
$\displaystyle\int {\dfrac{{{{\sec }^2}x/2}}{{\left( {\cos \lambda + 1} \right)\left( {1 - \cos x} \right){{\tan }^2}x/2}}}$
Let $\tan x/2 = k$
$\dfrac{1}{2}{\sec ^2}x/2 = dk$
$\displaystyle\int {\dfrac{2}{{\left( {\cos \lambda + 1} \right) - \left( {1 - \cos \lambda } \right) + 2}}dk}$
$\displaystyle\int {\dfrac{1}{{\sqrt {\cos \lambda + 1} - \sqrt {1 - \cos \lambda } k}} + \dfrac{1}{{\sqrt {\cos \lambda + 1} + \sqrt {1 - \cos \lambda } k}}}$
$= \dfrac{1}{{\sqrt {\cos \lambda + 1} }}\left[ {\dfrac{{ - \ln \sqrt {\cos \lambda + 1} - \sqrt {1 - \cos \lambda } k}}{{\sqrt {1 - \cos \lambda } }} + \dfrac{{\ln \sqrt {\cos \lambda + 1} - \sqrt {1 - \cos \lambda } k}}{{\sqrt {1 - \cos \lambda } }}} \right]$
$= \dfrac{1}{{\sqrt {1 - {{\cos }^2}\lambda } }}\,\ln \dfrac{{\sqrt {\cos \lambda + 1} + \sqrt {1 - \cos \lambda } k}}{{\sqrt {\cos \lambda + 1} - \sqrt {1 - \cos \lambda } k}} + C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Subjective Hard
Evaluate:
$\int { \cfrac { \sec ^{ 2 }{ x } }{ cosec ^{ 2 }{ x } } } dx$

1 Verified Answer | Published on 17th 09, 2020

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