Bonjour-besoin-d-aide-Calculer-ln-cosx-dx-

Question Number 113634 by eric last updated on 14/Sep/20

B o n j o u r b e s o i n d^{'} a i d e

C a l c u l e r \int l n (c o s x) d x

Answered by Olaf last updated on 14/Sep/20

\cos x = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k} x^{2 k}}{(2 k)!} \dots

Answered by MJS_new last updated on 18/Sep/20

\int \ln \cos x d x =

[by parts]

= x \ln \cos x + \int x \tan x d x

\int x \tan x d x = - i \int x \frac{e^{i x} - e^{- i x}}{e^{i x} + e^{i x}} d x =

= - i \int x \frac{e^{2 i x} - 1}{e^{2 i x} + 1} d x =

[t = e^{2 i x} \Leftrightarrow x = - \frac{i}{2} \ln t \to d x = - \frac{i}{2 t} d t]

= \frac{i}{4} \int \ln t \frac{t - 1}{t (t + 1)} d t =

\frac{i}{2} \int \frac{\ln t}{t + 1} d t - \frac{i}{4} \int \frac{\ln t}{t} d t

\frac{i}{2} \int \frac{\ln t}{t + 1} d t =

[by parts]

= \frac{i}{2} \ln t \ln (t + 1) - \frac{i}{2} \int \frac{\ln (t + 1)}{t} d t =

= \frac{i}{2} \ln t \ln (t + 1) + \frac{i}{2} {Li}_{2} (- t)

- \frac{i}{4} \int \frac{\ln t}{t} d t =

[by parts]

= \frac{i}{8} {(\ln t)}^{2}

\Rightarrow

\int x \tan x d x = \frac{i}{2} \ln t \ln (t + 1) + \frac{i}{2} {Li}_{2} (- t) + \frac{i}{8} {(\ln t)}^{2} =

= \frac{i}{2} \ln e^{2 i x} \ln (e^{2 i x} + 1) + \frac{i}{2} {Li}_{2} (- e^{2 i x}) + \frac{i}{8} {(\ln e^{2 i x})}^{2} =

= - \frac{i}{2} x^{2} - x \ln (e^{2 i x} + 1) + \frac{i}{2} L i_{2} (- e^{2 i x})

\Rightarrow

\int \ln \cos x d x =

= x \ln \cos x - \frac{i}{2} x^{2} - x \ln (e^{2 i x} + 1) + \frac{i}{2} L i_{2} (- e^{2 i x}) + C

hopefully I made no errors, please check!

Commented by Tawa11 last updated on 06/Sep/21

great sir

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