Question

$integration\frac{dx}{1 + \cos(2x) ?}$
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Answer 1

Topic :-

Indefinite Integration

To Integrate :-

$\displaystyle \int\dfrac{dx}{1+\cos 2x}$

Solution :-

$\displaystyle \int\dfrac{dx}{1+\cos 2x}$

$\displaystyle \int\dfrac{dx}{1+\cos^2x-\sin^2x}$

$(\because \cos 2x = cos^2x-sin^2x)$

$\displaystyle \int\dfrac{dx}{(1-\sin^2x)+\cos^2x}$

$\displaystyle \int\dfrac{dx}{\cos^2x+\cos^2x}$

$(\because 1-\sin^2x=\cos^2x)$

$\displaystyle \int\dfrac{dx}{2\cos^2x}$

$\displaystyle \int\dfrac{\sec^2x.dx}{2}$

$\left ( \because \dfrac{1}{\cos x}=\sec x\right )$

$\dfrac{1}{2}\displaystyle \int\sec^2x.dx$

$\dfrac{\tan x}{2}+C$

$\left (\because \displaystyle \int\sec^2x.dx=\tan x\right )$

$Here, C\:is\:constant\:of\:integration.$

Answer :-

$\dfrac{\tan x}{2}+C$

Additional Formulae :-

$\displaystyle \int \sin x = -\cos x+C$

$\displaystyle \int \cos x = \sin x+C$

$\displaystyle \int \tan x = \ln|\sec x|+C$

$\displaystyle \int \csc x = \ln|\csc x-\cot x|+C$

$\displaystyle \int \sec x = \ln|\sec x+\tan x|+C$

$\displaystyle \int \cot x = \ln|\sin x|+C$