cacuate-pi-4-pi-4-ln-1-a-cos-2-t-dt-with-a-lt-1-

Question Number 95844 by mathmax by abdo last updated on 28/May/20

cacuate \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (1 + a \cos^{2} t) dt with ∣ a ∣< 1

Answered by mathmax by abdo last updated on 29/May/20

let f (a) = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (1 + {acos}^{2} t) dt \Rightarrow f (a) = 2 \int_{0}^{\frac{π}{4}} \ln (1 + {acos}^{2} t) dt \Rightarrow

f^{^{'}} (a) = 2 \int_{0}^{\frac{π}{4}} \frac{\cos^{2} t}{1 + a \cos^{2} t} dt = \frac{2}{a} \int_{0}^{\frac{π}{4}} \frac{1 + {acos}^{2} t - 1}{1 + {acos}^{2} t} dt

= \frac{2}{a} \times \frac{π}{4} - \frac{2}{a} \int_{0}^{\frac{π}{4}} \frac{dt}{1 + {acos}^{2} t} = \frac{π}{2 a} - \frac{2}{a} \int_{0}^{\frac{π}{4}} \frac{dt}{1 + {acos}^{2} t} we have

\int_{0}^{\frac{π}{4}} \frac{dt}{1 + {acos}^{2} t} = \int_{0}^{\frac{π}{4}} \frac{dt}{1 + a \frac{1 + \cos (2 t)}{2}} = \int_{0}^{\frac{π}{4}} \frac{2 dt}{2 + a + acos (2 t)} =_{2 t = u}

= \int_{0}^{\frac{π}{2}} \frac{du}{2 + a + acosu} =_{\tan (\frac{u}{2}) = z} \int_{0}^{1} \frac{2 dz}{(1 + z^{2}) (2 + a + a \frac{1 - z^{2}}{1 + z^{2}})}

= \int_{0}^{1} \frac{2 dz}{(2 + a) (1 + z^{2}) + a - {az}^{2}} = \int_{0}^{1} \frac{2 dz}{2 + a + (2 + a - a) z^{2}} = \int_{0}^{1} \frac{2 dz}{2 + a + {2 z}^{2}}

= \frac{2}{2 + a} \int_{0}^{1} \frac{dz}{1 + \frac{2}{2 + a} z^{2}} =_{\sqrt{\frac{2}{2 + a}} z = α} \frac{2}{2 + a} \int_{0}^{\sqrt{\frac{2}{2 + a}}} \frac{\sqrt{\frac{2 + a}{2}} d α}{1 + α^{2}}

= \frac{2}{\sqrt{2}} \times \frac{1}{\sqrt{2 + a}} {[\arctan (α)]}_{0}^{\sqrt{\frac{2}{2 + a}}} = \frac{\sqrt{2}}{\sqrt{2 + a}} \arctan (\sqrt{\frac{2}{2 + a}}) \Rightarrow

f^{^{'}} (a) = \frac{π}{2 a} - \frac{2}{a} \times \frac{\sqrt{2}}{\sqrt{2 + a}} \arctan (\sqrt{\frac{2}{2 + a}}) \Rightarrow

f (a) = \frac{π}{2} \ln ∣ a ∣ - 2 \int \frac{1}{a} \sqrt{\frac{2}{2 + a}} \arctan (\sqrt{\frac{2}{2 + a}}) da + c

changement \sqrt{\frac{2}{2 + a}} = t give \frac{2}{2 + a} = t^{2} \Rightarrow \frac{2 + a}{2} = \frac{1}{t^{2}} \Rightarrow 2 + a = \frac{2}{t^{2}} \Rightarrow a = \frac{2}{t^{2}} - 2 \Rightarrow

da = 2 (\frac{- 2 t}{t^{4}}) = \frac{- 4}{t^{3}} \Rightarrow

\int \frac{1}{a} \sqrt{\frac{2}{2 + a}} \arctan (\sqrt{\frac{2}{2 + a}}) da = \int \frac{1}{\frac{2}{t^{2}} - 2} \times t \times \arctan (t) \times (\frac{- 4}{t^{3}}) dt

= - 4 \int \frac{1}{t^{2} (\frac{2}{t^{2}} - 2)} \arctan (t) dt = - 4 \int \frac{\arctan (t)}{2 - {2 t}^{2}} dt

= 2 \int \frac{arctant}{t^{2} - 1} dt = \int (\frac{1}{t - 1} - \frac{1}{t + 1}) \arctan (t) dt

= \int \frac{arctant}{t - 1} dt - \int \frac{arctant}{t + 1} dt (\to t = - u)

= \int \frac{arctant}{t - 1} dt - \int \frac{- arctanu}{- u + 1} (- du) = \int \frac{\arctan (t)}{t - 1} dt + \int \frac{arctanu}{u - 1} du

= 2 \int \frac{\arctan (t)}{t - 1} dt \dots . be continued \dots .

Commented by maths mind last updated on 21/Jun/20

\int \frac{a r c t a n (x)}{x - 1} d x = a r c t a n (x) l n ∣ x - 1 ∣ - \int \frac{l n (x - 1)}{x^{(2} + 1} d x

\int \frac{l n (x - 1)}{x^{2} + 1} d x = \frac{1}{2 i} \int \frac{l n (x - 1) (x + i - (x - i))}{(x + i) (x - i)} d x

= \frac{1}{2 i} \int \frac{l n (x - 1)}{x - i} - \frac{1}{2 i} \int \frac{l n (x - 1)}{(x + i)} d x

j = x - i \Rightarrow d j = d x

\int \frac{l n (j + i - 1)}{j} d j = \int \frac{l n (i - 1) l n (1 - \frac{j}{1 - i})}{\frac{j}{1 - i}} d (\frac{j}{1 - i})

= - l n (i - 1) {L i}_{2} (\frac{j}{1 - i})

\int \frac{l n (x - 1)}{x + i} d x = \int \frac{l n (y - i - 1)}{y} d y = \int \frac{l n (- i - 1) l n (1 - \frac{y}{i + 1})}{\frac{y}{i + 1}} . d (\frac{y}{1 + i})

= - l n (- 1 - i) {L i}_{2} (\frac{y}{i + 1})

w e g e t - \frac{1}{2 i} {l n (i - 1) {L i}_{2} (\frac{x - i}{1 - i}) - l n (- 1 - i) {L i}_{2} (\frac{x + i}{i + 1})}

Answered by mathmax by abdo last updated on 29/May/20

let determine f (a) = \int_{- \frac{π}{4}}^{\frac{π}{4}} \ln (1 + {acos}^{2} x) dx at form of serie

f (a) = 2 \int_{0}^{\frac{π}{4}} \ln (1 + a \times \frac{1 + \cos (2 x)}{2}) dx = 2 \int_{0}^{\frac{π}{4}} \ln (2 + a + acos (2 x)) dx

- 2 \times \frac{π}{4} \ln (2) + 2 \int_{0}^{\frac{π}{4}} \ln {(2 + a) (1 + \frac{a}{2 + a} \cos (2 x)} dx

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

we have \ln^{^{'}} (1 + u) = \frac{1}{1 + u} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{n} \Rightarrow \ln (1 + u) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} u^{n + 1}}{n + 1} + c (c = 0)

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow \int_{0}^{\frac{π}{4}} \ln (1 + \frac{a}{a + 2} \cos (2 x)) dx

=_{2 x = t} \int_{0}^{\frac{π}{2}} \ln (1 + \frac{a}{a + 2} cost) \frac{dt}{2} = \frac{1}{2} \int_{0}^{\frac{π}{2}} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {(\frac{a}{a + 2} cost)}^{n}) dt

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {(\frac{a}{a + 2})}^{n} \int_{0}^{\frac{π}{2}} \cos^{n} t dt = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} w_{n} {(\frac{a}{a + 2})}^{n}

with w_{n} = \int_{0}^{\frac{π}{2}} \cos^{n} t dt (wsllis integral) \Rightarrow

f (a) = \frac{π}{2} (\ln (2 + a) - \ln 2) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} w_{n} {(\frac{a}{a + 2})}^{n}

\dots . be continued \dots