0-pi-2-log-sin-x-2-log-cos-x-2-2-cos-pi-4-x-2-dx-

Question Number 139399 by mathsuji last updated on 26/Apr/21

\underset{0}{\int^{π / 2}} \frac{l o g (s i n (x / 2)) + l o g (c o s (x / 2))}{\sqrt{2} \cdot c o s (π / 4 - x / 2)} d x

Answered by Ar Brandon last updated on 26/Apr/21

Let u = \frac{π}{2} - x

Answered by Ar Brandon last updated on 26/Apr/21

I = \int_{0}^{\frac{π}{2}} \frac{\ln (\sin \frac{x}{2}) + \ln (\cos \frac{x}{2})}{\sqrt{2} \cos (\frac{π}{4} - \frac{x}{2})} d x

= \int_{0}^{\frac{π}{2}} \frac{\ln (\sin \frac{x}{2} \cos \frac{x}{2})}{\sqrt{2} \cos (\frac{1}{2} (\frac{π}{2} - x))} d x = \int_{0}^{\frac{π}{2}} \frac{\ln (\frac{1}{2}) + \ln (\sin x)}{\sqrt{2} \cos (\frac{1}{2} (\frac{π}{2} - x))} d x

= - \frac{\ln 2}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{d x}{\cos (\frac{1}{2} (\frac{π}{2} - x))} + \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{\ln (\sin x)}{\cos (\frac{1}{2} (\frac{π}{2} - x))} d x

= \sqrt{2} \ln 2 [\ln ∣ \sec (\frac{π}{4} - \frac{x}{2}) + \tan (\frac{π}{4} - \frac{x}{2}) ∣] + \frac{1}{\sqrt{2}} J

J = \int_{0}^{\frac{π}{2}} \frac{\ln (\cos x)}{\cos (\frac{x}{2})} d x

Answered by mathmax by abdo last updated on 26/Apr/21

I = \int_{0}^{\frac{π}{2}} \frac{\log (\sin (\frac{x}{2})) + \log (\cos (\frac{x}{2}))}{\sqrt{2} \cos (\frac{π}{4} - \frac{x}{2})} dx \Rightarrow

I = \int_{0}^{\frac{π}{2}} \frac{\log (\frac{sinx}{2})}{\sqrt{2} \cos (\frac{π}{4} - \frac{x}{2})} dx =_{\frac{π}{4} - \frac{x}{2} = t} \int_{\frac{π}{4}}^{o} \frac{\log (\frac{1}{2} \sin (\frac{π}{2} - 2 t))}{\sqrt{2} cost} (- 2) dt

= \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{\log (\cos (2 t)) - \log 2}{cost} dt

= \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{\log (\cos (2 t))}{cost} dt - \sqrt{2} \log (2) \int_{0}^{\frac{π}{4}} \frac{dt}{cost}

\int_{0}^{\frac{π}{4}} \frac{\log (\cos (2 t))}{cost} dt = \int_{0}^{\frac{π}{4}} \frac{\log ({2 \cos}^{2} t - 1)}{cost} dt

=_{cost = x} \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\log ({2 x}^{2} - 1)}{x} (- \frac{dx}{\sqrt{1 - x^{2}}})

= \int_{\frac{1}{\sqrt{2}}}^{1} \frac{\log ({2 x}^{2} - 1)}{x \sqrt{1 - x^{2}}} dx let f (a) = \int_{\frac{1}{\sqrt{2}}}^{1} \frac{\log ({2 x}^{2} - a)}{x \sqrt{1 - x^{2}}} dx with o < a < \sqrt{2}

f^{^{'}} (a) = - \int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{({2 x}^{2} - a) x \sqrt{1 - x^{2}}} dx

=_{x = \sin θ} - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\cos θ d θ}{({2 \sin}^{2} θ - 1) \sin θ . \cos θ}

= - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d θ}{\sin θ ({2 \sin}^{2} θ - a)}

F (u) = \frac{1}{u ({2 u}^{2} - 1)} = \frac{1}{u (\sqrt{2} u - \sqrt{a}) (\sqrt{2} u + \sqrt{a})} = \frac{m}{u} + \frac{b}{u \sqrt{2} - \sqrt{a}} + \frac{c}{u \sqrt{2} + \sqrt{a}}

\Rightarrow f^{^{'}} (a) = - m \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d θ}{\sin θ} - b \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d θ}{\sqrt{2} \sin θ - \sqrt{a}} - c \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d θ}{\sqrt{2} \sin θ + \sqrt{a}}

\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d θ}{\sin θ} =_{\tan (\frac{θ}{2}) = t} \int_{\sqrt{2} - 1}^{1} \frac{2 dt}{(1 + t^{2}) \times \frac{2 t}{1 + t^{2}}} = \int_{\sqrt{2} - 1}^{1} \frac{dt}{t}

= - \log (\sqrt{2} - 1)

\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{d θ}{\sqrt{2} \sin θ - \sqrt{a}} =_{\tan (\frac{θ}{2}) = y} \int_{\sqrt{2} - 1}^{1} \frac{2 dy}{(1 + y^{2}) (\sqrt{2} \frac{2 y}{1 + y^{2}} - \sqrt{a})}

= 2 \int_{\sqrt{2} - 1}^{1} \frac{dy}{2 \sqrt{2} y - \sqrt{a} - \sqrt{a} y^{2}} = - 2 \int_{\sqrt{2} - 1}^{1} \frac{dy}{\sqrt{a} y^{2} - 2 \sqrt{2} y + \sqrt{a}}

Δ^{^{'}} = 2 - a^{2} > 0 \Rightarrow y_{1} = \frac{\sqrt{2} + \sqrt{2 - a^{2}}}{\sqrt{a}} and y_{2} = \frac{\sqrt{2} - \sqrt{2 - a^{2}}}{\sqrt{a}}

\dots . be continued \dots .

Answered by qaz last updated on 27/Apr/21

\int_{0}^{π / 2} \frac{l n (\sin \frac{x}{2}) + l n (\cos \frac{x}{2})}{\sqrt{2} \cos (\frac{π}{4} - \frac{x}{2})} d x

= 2 \int_{0}^{π / 4} \frac{l n \sin x + l n \cos x}{\cos x + \sin x} d x

= 2 \int_{0}^{π / 4} \frac{\cos x - \sin x}{\cos 2 x} (l n \sin x + l n \cos x) d x

= 2 \int_{0}^{π / 4} \frac{\cos x}{\cos 2 x} (l n \sin x + l n \cos x) d x - 2 \int_{0}^{π / 4} \frac{\sin x}{\cos 2 x} (l n \sin x + l n \cos x) d x

= 2 \int_{0}^{π / 4} \frac{l n \cos x}{\cos 2 x} \cos x d x - 2 \int_{0}^{π / 4} \frac{l n \sin x}{\cos 2 x} \sin x d x

= \int_{0}^{π / 4} \frac{l n (1 - \sin^{2} x)}{1 - 2 \sin^{2} x} d (\sin x) + \int_{0}^{π / 4} \frac{l n (1 - \cos^{2} x)}{2 \cos^{2} x - 1} d (\cos x)

= \int_{0}^{1 / \sqrt{2}} \frac{l n (1 - u^{2})}{1 - 2 u^{2}} d u + \int_{0}^{1 / \sqrt{2}} \frac{l n (1 - v^{2})}{2 v^{2} - 1} d v

= 0

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