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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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