0-pi-4-ln-1-2-cos-x-dx-

Question Number 161660 by amin96 last updated on 20/Dec/21

\int_{0}^{\frac{π}{4}} l n (1 + \sqrt{2} c o s (x)) d x = ? ? ?

Answered by mindispower last updated on 21/Dec/21

\int_{0}^{\frac{π}{4} r}

l n (1 + \sqrt{2} c o s (\frac{π}{4} - x)) d x 3

= l n (1 + c o s (x) + s i n (x))) d x

= \int_{0}^{\frac{π}{4}} l n (2 {c o s}^{2} (\frac{x}{2}) + 2 s i n (\frac{x}{2}) c o s (\frac{x}{2})) d x

= (\frac{π}{4} l n (2) + \int_{0}^{\frac{π}{4}} l n (c o s (\frac{x}{2})) d x + n \int_{0}^{\frac{π}{4}} l n (\sqrt{2} s n i n (\frac{x}{2} + \frac{π}{4}))

= \frac{3 π l n (2)}{8} + 2 \int_{0}^{\frac{π}{8}} l n (c o s (t)) + 2 \int_{0}^{\frac{π}{8}} l n (s i n (t + \frac{π}{4})) d t

=, \frac{3 π l n (2)}{8} 4 + 2 \int_{\frac{π}{4}}^{\frac{3 π}{8}} l n (s i n (u)) d, Σ

l n (s i n (x)) = - l n (2) - \sum_{n ⩾ 1} \frac{c o s (2 n x)}{n}

w e g e t \frac{3 π l n (2)}{8} - \frac{π}{4} l n (2) - \sum_{n ⩾ 1} \frac{s i n (n 3 \frac{π}{4}) - s i n (\frac{n π}{2})}{n^{2}}

= \frac{π l n (2)}{8} - {C l}_{2} (\frac{3 π}{4}) + {C l}_{2} (\frac{π}{2}),

{C l}_{2} (z) = \sum_{n ⩾ 1} \frac{s i n (n z)}{n^{2}}, C l a u s e n f u n c t i o n

w e c a n e x p r e s s w i t h e e l e m e n t r y f u n c t i o n

Commented by Ar Brandon last updated on 21/Dec/21

g \overset{´}{e n i a l}!

Answered by mathmax by abdo last updated on 21/Dec/21

f (a) = \int_{0}^{\frac{π}{4}} \ln (1 + acosx) dx (a > 1)

f^{^{'}} (a) = \int_{0}^{\frac{π}{4}} \frac{cosx}{1 + acosx} dx = \frac{1}{a} \int_{0}^{\frac{π}{4}} \frac{1 + acosx - 1}{1 + acosx} dx

= \frac{π}{4 a} - \frac{1}{a} \int_{0}^{\frac{π}{4}} \frac{dx}{1 + acosx} changement \tan (\frac{x}{2}) = t give

\int_{0}^{\frac{π}{4}} \frac{dx}{1 + acosx} = \int_{0}^{\sqrt{2} - 1} \frac{2 dt}{(1 + t^{2}) (1 + a \frac{1 - t^{2}}{1 + t^{2}})}

= 2 \int_{0}^{\sqrt{2} - 1} \frac{dt}{1 + t^{2} + a - {at}^{2}} = 2 \int_{0}^{\sqrt{2} - 1} \frac{dt}{(1 - a) t^{2} + 1 + a}

= - 2 \int_{0}^{\sqrt{2} - 1} \frac{dt}{(a - 1) t^{2} - (1 + a)} = \frac{- 2}{a - 1} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t^{2} - \frac{a + 1}{a - 1}}

=_{t = \sqrt{\frac{a + 1}{a - 1}} u} \frac{- 2}{a - 1} \int_{0}^{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}}} \frac{\sqrt{\frac{a + 1}{a - 1}}}{\frac{a + 1}{a - 1} (u^{2} - 1)} du

= \frac{a - 1}{a + 1} \times \frac{\sqrt{a + 1}}{\sqrt{a - 1}} \times \frac{- 2}{a - 1} \int_{0}^{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}}} \frac{du}{u^{2} - 1}

= \frac{\sqrt{a - 1}}{\sqrt{a + 1}} \times \frac{- 2}{a - 1} \int \dots du

= - \frac{1}{\sqrt{a^{2} - 1}} \int_{0}^{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}}} (\frac{1}{u - 1} - \frac{1}{u + 1}) du

= \frac{1}{\sqrt{a^{2} - 1}} {[\ln ∣ \frac{u + 1}{u - 1} ∣]}_{0}^{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}}}

= \frac{1}{\sqrt{a^{2} - 1}} {\ln ∣ \frac{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}} + 1}{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}} - 1} ∣} \Rightarrow

f^{^{'}} (a) = \frac{π}{4 a} - \frac{1}{a \sqrt{a^{2} - 1}} \ln ∣ \frac{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}} + 1}{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}} - 1} ∣ \Rightarrow

f (a) = \frac{π}{4} lna - \int \frac{1}{a \sqrt{a^{2} - 1}} \ln (\frac{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}} + 1}{(\sqrt{2} - 1) \sqrt{\frac{a - 1}{a + 1}} - 1}) + C

\dots be continued \dots

Answered by Lordose last updated on 23/Dec/21

Ω = \int_{0}^{\frac{π}{4}} \ln (1 + \sqrt{2} \cos (x)) dx = \int_{0}^{\frac{π}{4}} \ln (1 + \cos (x) + \sin (x)) dx

Ω = \int_{0}^{\frac{π}{4}} \ln ({2 \cos}^{2} (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})) dx

Ω = \ln (2) \int_{0}^{\frac{π}{4}} 1 dx + \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{x}{2})) dx + \int_{0}^{\frac{π}{4}} \ln (\sin (\frac{x}{2}) + \cos (\frac{x}{2})) dx

Ω = \frac{π \ln (2)}{4} + \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{x}{2})) dx + \frac{1}{2} \ln (2) \int_{0}^{\frac{π}{4}} 1 dx + \int_{0}^{\frac{π}{4}} \ln (\sin (\frac{x}{2} + \frac{π}{4})) dx

N . B :: \int_{0}^{x} \ln (\sin (\frac{t}{2})) dt = - {Cl}_{2} (x) - xln (2)

{Cl}_{2} (π) = 0; {Cl}_{2} (\frac{π}{2}) = G ({Catalan}^{'} s Constant)

Ω = \frac{3 π \ln (2)}{8} + A + B

A \overset{\frac{x}{2} = \frac{π}{2} - \frac{u}{2}}{=} \int_{\frac{3 π}{4}}^{π} \ln (\sin (\frac{u}{2})) du = \int_{0}^{π} \ln (\sin (\frac{u}{2})) du - \int_{0}^{\frac{3 π}{4}} \ln (\sin (\frac{u}{2})) du

A = - {Cl}_{2} (π) - π \ln (2) + {Cl}_{2} (\frac{3 π}{4}) + \frac{3 π \ln (2)}{4}

A = {Cl}_{2} (\frac{3 π}{4}) - \frac{π \ln (2)}{4}

B \overset{\frac{x}{2} = \frac{u}{2} - \frac{π}{4}}{=} \int_{\frac{π}{2}}^{\frac{3 π}{4}} \ln (\sin (\frac{u}{2})) du = \int_{0}^{\frac{3 π}{4}} \ln (\sin (\frac{u}{2})) du - \int_{0}^{\frac{π}{2}} \ln (\sin (\frac{u}{2})) du

B = - {Cl}_{2} (\frac{3 π}{4}) - \frac{3 π \ln (2)}{4} + {Cl}_{2} (\frac{π}{2}) + \frac{π \ln (2)}{2}

B = {Cl}_{2} (\frac{π}{2}) - {Cl}_{2} (\frac{3 π}{4}) - \frac{π \ln (2)}{4}

Ω = \frac{3 π \ln (2)}{8} + A + B

Ω = \frac{3 π \ln (2)}{8} + {Cl}_{2} (\frac{3 π}{4}) - \frac{π \ln (2)}{4} + {Cl}_{2} (\frac{π}{2}) - {Cl}_{2} (\frac{3 π}{4}) - \frac{π \ln (2)}{4}

Ω = {Cl}_{2} (\frac{π}{2}) - \frac{π \ln (2)}{8}

Ω = G - \frac{π \log (2)}{8}

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