explicit-g-a-0-pi-4-ln-1-acos-2-d-

Question Number 113630 by mathmax by abdo last updated on 14/Sep/20

explicit g (a) = \int_{0}^{\frac{π}{4}} \ln (1 + {acos}^{2} θ) d θ

Answered by Dwaipayan Shikari last updated on 15/Sep/20

I (a) = \int_{0}^{\frac{π}{4}} l o g (1 + {a c o s}^{2} θ) d θ

I^{'} (a) = \int_{0}^{\frac{π}{4}} \frac{{c o s}^{2} θ}{1 + {a c o s}^{2} θ}

I^{'} (a) = \frac{1}{a} \int_{0}^{\frac{π}{4}} 1 - \frac{1}{1 + {a c o s}^{2} θ}

I^{^{'}} (a) = \frac{π}{4 a} - \int^{\frac{π}{4}} \frac{{s e c}^{2} θ}{{s e c}^{2} θ + a} d θ

I^{'} (a) = \frac{π}{4 a} - \int_{0}^{\frac{π}{4}} \frac{{s e c}^{2} θ}{{t a n}^{2} θ + 1 + a}

I^{'} (a) = \frac{π}{4 a} - \int_{0}^{1} \frac{d t}{t^{2} + {(\sqrt{1 + a})}^{2}}

I^{'} (a) = \frac{π}{4 a} - {[\frac{1}{\sqrt{1 + a}} {t a n}^{- 1} \frac{t}{\sqrt{1 + a}}]}_{0}^{1}

I (a) = \int \frac{π}{4 a} - \int \frac{1}{\sqrt{1 + a}} {t a n}^{- 1} \frac{1}{\sqrt{1 + a}} d a

I (a) = \frac{π}{4} l o g (a) - \int \frac{2 u}{u} . {t a n}^{- 1} \frac{1}{u} d u 1 + a = u^{2}, 1 = 2 u \frac{d u}{d a}

I (a) = \frac{π}{4} l o g (a) - 2 \int {t a n}^{- 1} \frac{1}{u} d u {t a n}^{- 1} \frac{1}{u} = α

I (a) = \frac{π}{4} l o g (a) - 2 {u t a n}^{- 1} \frac{1}{u} - 2 \int \frac{1}{u \sqrt{1 + \frac{1}{u^{2}}}} d u

I (a) = \frac{π}{4} l o g (a) - 2 \sqrt{1 + a} {t a n}^{- 1} \frac{1}{\sqrt{1 + a}} - 2 l o g (u + \sqrt{u^{2} + 1}) + C

I (a) = \frac{π}{4} l o g (a) - 2 \sqrt{1 + a} {t a n}^{- 1} \frac{1}{\sqrt{1 + a}} - 2 l o g (\sqrt{1 + a} + \sqrt{2 + a}) + C

I (- 1) = \frac{π^{2}}{4} i + C = \int_{0}^{\frac{π}{4}} l o g (1 - {c o s}^{2} x) d x

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

= - \frac{π}{2} l o g (2) + \int_{- \frac{π}{2}}^{0} l o g (- 1) + l o g (c o s x) = \frac{π^{2}}{2} i

I (- 1) = \frac{π^{2}}{4} i + C = \frac{π^{2} i}{2} \Rightarrow C = \frac{π^{2}}{4} i

I (a) = \frac{π}{4} l o g (a) - 2 \sqrt{1 + a} {t a n}^{- 1} \frac{1}{\sqrt{1 + a}} - 2 l o g (\sqrt{1 + a} + \sqrt{2 + a}) + \frac{π^{2}}{4} i

Commented by Tawa11 last updated on 06/Sep/21

great sir

Answered by mathmax by abdo last updated on 14/Sep/20

∣ a ∣< 1

Answered by Olaf last updated on 14/Sep/20

\frac{1}{1 + a \cos^{2} θ} = \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} a^{k} \cos^{2 k} θ

- \frac{2 a \cos θ \sin θ}{1 + a \cos^{2} θ} = - 2 a \sin θ \cos θ \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} a^{k} \cos^{2 k} θ

- \frac{2 a \cos θ \sin θ}{1 + a \cos^{2} θ} = - 2 a \sin θ \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} a^{k} \cos^{2 k + 1} θ

\ln (1 + a \cos^{2} θ) = - 2 a \underset{k = 0}{\sum^{\infty}} {(- 1)}^{k} a^{k} \frac{\cos^{2 k + 2} θ}{2 k + 2}

\int_{0}^{\frac{π}{4}} \ln (1 + a \cos^{2} θ) d θ = - 2 a \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k} a^{k}}{2 k + 2} \int_{0}^{\frac{π}{4}} \cos^{2 k + 2} θ d θ

\cos^{2 k} θ = {(\frac{1 + \cos 2 θ}{2})}^{k}

\cos^{2 k} θ = \frac{1}{2^{k}} \underset{p = 0}{\sum^{k}} C_{k}^{p} \cos^{p} 2 θ

\int_{0}^{\frac{π}{4}} \ln (1 + a \cos^{2} θ) d θ = - 2 a \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k} a^{k}}{2 k + 2} {[\frac{\cos^{2} θ}{2^{k}} \underset{p = 0}{\sum^{k}} C_{k}^{p} \cos^{p} 2 θ]}_{0}^{\frac{π}{4}}

\int_{0}^{\frac{π}{4}} \ln (1 + a \cos^{2} θ) d θ = 2 a \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k} a^{k}}{(2 k + 2) 2^{k}} \underset{p = 0}{\sum^{k}} C_{k}^{p}

\int_{0}^{\frac{π}{4}} \ln (1 + a \cos^{2} θ) d θ = 2 a \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k} a^{k}}{(2 k + 2)}

\dots

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