evaluate-0-pi-log-a-cos-x-dx-

Question Number 171198 by infinityaction last updated on 09/Jun/22

e v a l u a t e

\int_{0}^{π} \log (a + \cos x) d x

Answered by aleks041103 last updated on 09/Jun/22

I (a) = \int_{0}^{π} l n (a + c o s x) d x

I^{'} (a) = \int_{0}^{π} \frac{d x}{a + c o s (x)}

t = t g (x / 2)

\Rightarrow c o s x = 1 - 2 {s i n}^{2} (x / 2) =

= 1 - \frac{2}{{c s c}^{2} (x / 2)} = 1 - \frac{2}{1 + {c t g}^{2} (x / 2)} =

= 1 - \frac{2}{1 + \frac{1}{{t g}^{2} (x / 2)}} = 1 - \frac{2}{1 + \frac{1}{t^{2}}} =

= 1 - \frac{2 t^{2}}{1 + t^{2}} = \frac{1 - t^{2}}{1 + t^{2}}

x = 2 a r c t g (t) \Rightarrow d x = \frac{2 d t}{1 + t^{2}}

x = 0 \to t = 0

x = π \to t \to \infty

\Rightarrow I^{'} (a) = \int_{0}^{\infty} \frac{2 d t}{a (1 + t^{2}) + 1 - t^{2}} =

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

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

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

= \frac{2}{\sqrt{a^{2} - 1}} \frac{π}{2} = \frac{π}{\sqrt{a^{2} - 1}} = I^{'} (a)

\Rightarrow I (a) = I (1) + π \int_{1}^{a} \frac{d a}{\sqrt{a^{2} - 1}}

a = c o s h (x)

d a = s i n h (x) d x

\Rightarrow \int \frac{d a}{\sqrt{a^{2} - 1}} = \int \frac{s i n h (x) d x}{\sqrt{{c o s h}^{2} x - 1}} = x = a r c c o s h (a)

\Rightarrow I (a) = I (1) + π (a r c c o s h (a) - a r c c o s h (1))

I (a) = I (1) + π a r c c o s h (a)

I (1) = \int_{0}^{π} l n (1 + c o s (x)) d x

l n (1 + c o s (x)) = l n (2 {c o s}^{2} (x / 2)) =

= l n (2) + 2 l n (c o s (x / 2))

\Rightarrow I (1) = π l n (2) + 4 \int_{0}^{π / 2} l n (c o s (x)) d x

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

2 i = \int_{0}^{π / 2} l n (s i n x c o s x) d x = \int_{0}^{π / 2} l n (\frac{1}{2} s i n (2 x)) d x =

= - \frac{π l n (2)}{2} + \int_{0}^{π / 2} l n (s i n (2 x)) d x =

= - \frac{π l n (2)}{2} + \frac{1}{2} \int_{0}^{π} l n (s i n (u)) d u =

= - \frac{π l n (2)}{2} + i = 2 i

\Rightarrow i = - \frac{π l n (2)}{2}

\Rightarrow I (1) = π l n (2) + 4 (- \frac{π l n (2)}{2}) = - π l n (2)

\Rightarrow I (a) = π (a r c c o s h (a) - l n (2))

c o s h (x) = \frac{e^{x} + e^{- x}}{2} = \frac{t + \frac{1}{t}}{2} = k

\Rightarrow 2 k t = t^{2} + 1

\Rightarrow t^{2} - 2 k t + 1 = 0

\Rightarrow t = e^{x} = \frac{2 k \pm \sqrt{4 k^{2} - 4}}{2} = k \pm \sqrt{k^{2} - 1}

\Rightarrow a r c c o s h (x) = l n (x + \sqrt{x^{2} - 1})

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

Commented by infinityaction last updated on 10/Jun/22

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