If sec 2 x - 1 ? d x = log e | cos 2 x + cos 2 x ( 1 + cos 1 x ) | + constant, then is equal to _________ . [2023]

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Solution

Q.
If $\int \sqrt{\sec 2 x - 1} d x = α \log_{e}$ $| \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} |$ $+$ constant, then $β - α$ is equal to _________ . [2023]

Ans.
(1)

$Let I = \int \sqrt{(\sec 2 x - 1)} d x = \int \sqrt{\frac{1 - \cos 2 x}{\cos 2 x}} d x$

$= \int \sqrt{\frac{2 \sin^{2} x}{2 \cos^{2} x - 1}} d x$ $= \sqrt{2} \int \frac{\sin x}{\sqrt{2 \cos^{2} x - 1}} d x$

$Substitute \cos x = t \Rightarrow - \sin x d x = d t$

$I = - \sqrt{2} \int \frac{d t}{\sqrt{2 t^{2} - 1}} = - \ln$ $| \sqrt{2} t + \sqrt{2 t^{2} - 1} |$

$= - \ln$ $| \sqrt{2} \cos x + \sqrt{2 \cos^{2} x - 1} |$

$= - \ln$ $| \sqrt{2} \cos x + \sqrt{\cos 2 x} |$

$= - \frac{1}{2} \ln$ $| 2 \cos^{2} x + \cos 2 x + 2 \sqrt{\cos 2 x} \cdot \sqrt{2} \cos x |$ $+ C$

$= \frac{- 1}{2} \ln (2 \cos 2 x + 1 + 2 \sqrt{\cos 2 x} \sqrt{1 + \cos 2 x})$

$= \frac{- 1}{2} \ln (\cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x} \sqrt{1 + \cos 2 x})$

$= \frac{- 1}{2} \ln$ $| \cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x} \cdot \sqrt{1 + \cos 2 x} |$ $+ C$

$\Rightarrow - \frac{1}{2} \ln$ $| \cos x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)} |$ $+ C$

$= α \ln$ $| \cos 2 x + β |$ $+ \sqrt{\cos 2 x (1 + \cos \frac{1}{β})} + c$

On comparing, we get, $α = - \frac{1}{2}, β = \frac{1}{2}$ $∴$ $β - α = \frac{1}{2} + \frac{1}{2} = 1$

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