show that ∫_0 ^(π/4) ((ln(2cos^2 x))/(cos2x))dx=(π^2 /(16))


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Question Number 111617 by mathdave last updated on 04/Sep/20

$s h o w t h a t$ $\int_{0}^{\frac{π}{4}} \frac{\ln ({2 \cos}^{2} x)}{\cos 2 x} d x = \frac{π^{2}}{16}$
	Answered by mathmax by abdo last updated on 04/Sep/20

	$A = \int_{0}^{\frac{π}{4}} \frac{\ln ({2 \cos}^{2} x)}{\cos (2 x)} dx \Rightarrow A = \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \cos (2 x))}{\cos (2 x)} dx$ $=_{2 x = t} \int_{0}^{\frac{π}{2}} \frac{\ln (1 + cost)}{cost} dt let f (a) = \int_{0}^{\frac{π}{2}} \frac{\ln (1 + acost)}{cost} dt (o < a < 1)$ $we have f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{cost}{(1 + acost) cost} dt = \int_{0}^{\frac{π}{2}} \frac{dt}{1 + acost}$ $=_{\tan (\frac{t}{2}) = u} \int_{0}^{1} \frac{2 du}{(1 + u^{2}) (1 + a \frac{1 - u^{2}}{1 + u^{2}})} = 2 \int_{0}^{1} \frac{du}{1 + u^{2} + a - {au}^{2}}$ $= 2 \int_{0}^{1} \frac{du}{(1 - a) u^{2} + 1 + a} = \frac{2}{1 - a} \int_{0}^{1} \frac{du}{u^{2} + \frac{1 + a}{1 - a}}$ $=_{u = \sqrt{\frac{1 + a}{1 - a}} z} \frac{2}{1 - a} \times \frac{1 - a}{1 + a} \int_{0}^{\sqrt{\frac{1 - a}{1 + a}}} \frac{1}{1 + z^{2}} \times \frac{\sqrt{1 + a}}{\sqrt{1 - a}} dz$ $= \frac{2}{\sqrt{1 - a^{2}}} \arctan \sqrt{\frac{1 - a}{1 + a}} \Rightarrow$ $f (a) = 2 \int_{0}^{a} \frac{\arctan \sqrt{\frac{1 - t}{1 + t}}}{\sqrt{1 - t^{2}}} dt + c (c = f (0) = 0) \Rightarrow$ $f (a) = 2 \int_{0}^{a} \frac{\arctan \sqrt{\frac{1 - t}{1 + t}}}{\sqrt{1 - t^{2}}} dt and$ $\int_{0}^{\frac{π}{2}} \frac{\ln (1 + cosx)}{cosx} dx = \lim_{a \to 1} f (a) = 2 \int_{0}^{1} \frac{\arctan \sqrt{\frac{1 - t}{1 + t}}}{\sqrt{1 - t^{2}}} dt$ $=_{t = \cos θ} 2 \int_{\frac{π}{2}}^{0} \frac{\arctan \sqrt{\frac{{2 \sin}^{2} (\frac{θ}{2})}{{2 \cos}^{2} (\frac{θ}{2})}}}{\sin θ} (- \sin θ) d θ$ $= 2 \int_{0}^{\frac{π}{2}} \arctan (\tan (\frac{θ}{2})) d θ = 2 \int_{0}^{\frac{π}{2}} \frac{θ}{2} d θ = {[\frac{θ^{2}}{2}]}_{0}^{\frac{π}{2}} = \frac{π^{2}}{8} \Rightarrow$ $★ \int_{0}^{\frac{π}{4}} \frac{\ln ({2 \cos}^{2} x)}{\cos (2 x)} dx = \frac{π^{2}}{8} ★$
	Commented by mathmax by abdo last updated on 04/Sep/20

	$sorry error at line 2 we have A = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\ln (1 + cosx)}{cosx} dx \Rightarrow$ $A = \frac{π^{2}}{16}$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
	Answered by mathdave last updated on 04/Sep/20

	$s o l u t i o n$ $f r o m {2 \cos}^{2} x = 1 + \cos 2 x$ $I = \int_{0}^{\frac{π}{4}} \frac{\ln ({2 \cos}^{2} x)}{\cos 2 x} d x = \int_{0}^{\frac{π}{4}} \frac{\ln (1 + \cos 2 x)}{\cos 2 x} d x$ $l e t y = 2 x$ $I = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\ln (\cos y)}{\cos y} d y u s i n g D u i s$ $l e t I (α) = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\ln (1 + α \cos y)}{\cos y} d y$ $I^{^{'}} (α) = \frac{1}{2} \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{1}{1 + α \cos y} d α d y$ $b u t \cos y = \frac{1 - \tan^{2} \frac{y}{2}}{1 + \tan^{2} \frac{y}{2}} = \frac{1 - t^{2}}{1 + t^{2}} l e t t = \tan \frac{y}{2} a n d d y = \frac{2 d t}{1 + t^{2}}$ $I = \frac{1}{2} \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{1}{1 + α (\frac{1 - t^{2}}{1 + t^{2}})} \times \frac{2 d t}{1 + t^{2}} d t d α$ $I = \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{1}{(1 + t^{2} + α - α t^{2})} d t d α = \int_{0}^{\frac{π}{2}} \int_{0}^{1} \frac{1}{(1 + α) + (1 - α) t^{2}} d t d α$ $I (α) = \int_{0}^{\frac{π}{2}} \frac{1}{1 - α} \int_{0}^{1} \frac{1}{{(\sqrt{\frac{1 + α}{1 - α}})}^{2} + t^{2}} d α d t$ $b u t t = \tan \frac{y}{2} = \tan \frac{π}{4} = 1$ $I (α) = \int_{0}^{1} \frac{1}{1 - α} ∙ \frac{1}{\sqrt{\frac{1 + α}{1 - α}}} \tan^{- 1} {[\frac{t}{\sqrt{\frac{1 + α}{1 - α}}}]}_{0}^{\frac{π}{2}} d α$ $I (α) = \int_{0}^{1} \frac{1}{1 - α} ∙ \frac{\sqrt{1 - α^{2}}}{1 + α} \tan^{- 1} [\frac{1}{\sqrt{\frac{1 + α}{1 - α}}}] d α$ $I (α) = \int_{0}^{1} \frac{1}{\sqrt{1 - α^{2}}} \tan^{- 1} [\sqrt{\frac{1 - α}{1 + α}}] d α$ $l e t α = \cos y a n d d α = - \sin y d y$ $a n d \tan \frac{y}{2} = \sqrt{\frac{1 - \cos y}{1 + \cos y}} = \sqrt{\frac{1 - α}{1 + α}}$ $= \int_{\frac{π}{2}}^{0} \frac{1}{\sin y} \tan^{- 1} [\sqrt{\frac{1 - \cos y}{1 + \cos y}}] (- \sin y) d y$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} \tan^{- 1} [\tan \frac{y}{2}] d y = \int_{0}^{\frac{π}{2}} \frac{y}{2} d y = \frac{1}{2} {[\frac{y^{2}}{2}]}_{0}^{\frac{π}{2}}$ $= \frac{1}{4} [{(\frac{π}{2})}^{2}] = \frac{π^{2}}{16}$ $\int_{0}^{\frac{π}{4}} \frac{\ln ({2 \cos}^{2} x)}{\cos 2 x} d x = \frac{π^{2}}{16} Q . E . D$ $b y m a t h d a v e (04 / 09 / 2020)$
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