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
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