Question-45498

Question Number 45498 by Sanjarbek last updated on 13/Oct/18

Commented by Meritguide1234 last updated on 13/Oct/18

n o t s o l v a b l e

Commented by MJS last updated on 13/Oct/18

possible but not elementary . I^{'} ll post it later

Commented by maxmathsup by imad last updated on 13/Oct/18

l e t f (t) = \int l n (s i n (t x)) d x \Rightarrow f^{^{'}} (t) = \int \frac{x c o s (t x)}{s i n (t x)} d x

=_{t x = u} \int \frac{u}{t} \frac{c o s (u)}{s i n (u)} \frac{d u}{t} = \frac{1}{t^{2}} \int u \frac{c o s (u)}{s i n (u)} d u a l s o c h a n g e m e n t t a n (\frac{u}{2}) = α

g i v e f (t) = \frac{1}{t^{2}} \int 2 a r c t a n (α) \frac{\frac{1 - α^{2}}{1 + α^{2}}}{\frac{2 α}{1 + α^{2}}} \frac{2 d α}{1 + α^{2}}

= \frac{4}{t^{2}} \int \frac{a r c t a n (α)}{1 + α^{2}} d α l e t t h e n i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n

φ (u) = \int \frac{a r c t a n (u α)}{1 + α^{2}} d α \Rightarrow φ^{^{'}} (u) = \int \frac{α}{(1 + α^{2}) (1 + u^{2} α^{2})} d α l e t d e c o m p o s e

F (α) = \frac{a α + b}{1 + α^{2}} + \frac{c α + d}{1 + u^{2} α^{2}} \dots . b e c o n t i n u e d \dots

Answered by MJS last updated on 14/Oct/18

\int \ln \sin x d x = \int \ln (- \frac{i}{2} (e^{i x} - e^{- i x})) d x =

[t = i x \to d x = - i d t

= - i \int \ln (- \frac{i}{2} (e^{t} - e^{- t})) d t =

= - i \int \ln (- \frac{i}{2}) d t - i \int \ln (e^{t} - e^{- t}) d t

the first one is easy :

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

= \ln (- \frac{i}{2}) x

the second one :

- i \int \ln (e^{t} - e^{- t}) d t =

[\begin{matrix} \int f^{'} g = f g - \int {f g}^{'} \\ f^{'} = 1 \to f = t \\ g = \ln (e^{t} - e^{- t}) \to g^{'} = \frac{e^{t} + e^{- t}}{e^{t} - e^{- t}} \end{matrix}]

= - i t \ln (e^{t} - e^{- t}) + i \int t \frac{e^{t} + e^{- t}}{e^{t} - e^{- t}} d t

the first term :

- i t \ln (e^{t} - e^{- t}) = x \ln (e^{i x} - e^{- i x})

the second term :

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

[u = e^{2 t} - 1 \to d t = \frac{e^{- 2 t}}{2} d u]

= \frac{i}{4} \int \frac{(u + 2) \ln (u + 1)}{u (u + 1)} d u =

= \frac{i}{2} \int \frac{\ln (u + 1)}{u} d u + \frac{i}{4} \int \frac{\ln (u + 1)}{u + 1} d u

the first one :

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

[v = - u \to d u = - d v]

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

this is a special integral (dilogarithm)

= - \frac{i}{2} {Li}_{2} v = - \frac{i}{2} {Li}_{2} (- u) = - \frac{i}{2} {Li}_{2} (1 - e^{2 t}) =

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

the second one :

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

[w = \ln (u + 1) \to d u = (u + 1) d w]

= \frac{i}{4} \int w d w = \frac{i}{8} w^{2} = \frac{i}{8} \ln^{2} (u + 1) = \frac{i}{8} \ln^{2} (e^{2 t}) =

= \frac{i}{2} t^{2} = - \frac{i}{2} x^{2}

so we have

\int \ln \sin x d x =

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

please check \dots

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