0-pi-2-arcos-cosx-1-2cosx-dx-

Question Number 154437 by ArielVyny last updated on 18/Sep/21

\int_{0}^{\frac{π}{2}} a r c o s (\frac{c o s x}{1 + 2 c o s x}) d x

Answered by phanphuoc last updated on 18/Sep/21

p u t x = t a n (t / 2)

Answered by Kamel last updated on 18/Sep/21

\frac{5 π^{2}}{24}

Answered by puissant last updated on 18/Sep/21

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

= 2 \int_{0}^{\frac{π}{2}} a r c t a n (\sqrt{\frac{1 + c o s x}{1 + 3 c o s x}}) d x = 4 \int_{0}^{\frac{π}{4}} a r c t a n (\sqrt{\frac{1 + c o s 2 u}{1 + 3 c o s 2 u}}) d u

D i v e r :^{″} 1 + c o s 2 u = 2 c o s u e t c o s 2 u = 1 - {s i n}^{2} u {(T r i v i a l)}^{″}

\Rightarrow Q = 4 \int_{0}^{\frac{π}{4}} a r c t a n (\sqrt{\frac{{c o s}^{2} u}{1 - {s i n}^{2} u}}) d u = 4 \int_{0}^{\frac{π}{4}} a r c t a n (\frac{c o s u}{\sqrt{1 - 2 s i n u}}) d u

= 4 \int_{0}^{\frac{π}{4}} \int_{0}^{1} \frac{\sqrt{2 - 3 {s i n}^{2} u} c o s u}{(x^{2} + 2) - {s i n}^{2} u (x^{2} + 3)} d x d u

\sqrt{3} s i n u = \sqrt{2} s i n θ \to c o s u d u = \frac{\sqrt{2}}{\sqrt{3}} c o s θ d θ

\Rightarrow Q = 4 \int_{0}^{\frac{π}{3}} \int_{0}^{1} \frac{2 \sqrt{3} {c o s}^{2} θ}{3 x^{2} + 6 - 2 (1 - {c o s}^{2} θ) (x^{2} + 3)} d x d θ

\overset{t = t a n u}{=} 8 \sqrt{3} \int_{0}^{\sqrt{3}} \int_{0}^{1} \frac{1}{(1 + t^{2}) (t^{2} x^{2} + 3 x^{2} + 6)} d x d t

= 8 \sqrt{3} \int_{0}^{\sqrt{3}} \int_{0}^{1} (\frac{- \frac{1}{2 x^{2} + 6}}{1 + t^{2}} + \frac{\frac{x^{2}}{2 x^{2} + 6}}{t^{2} x^{2} + 3 x^{2} + 6}) d x d t = 8 \sqrt{3} \int_{0}^{1} \frac{1}{2 x^{2} + 6} (\int_{0}^{\sqrt{3}} \frac{d t}{1 + t^{2}} - \int_{0}^{\sqrt{3}} \frac{d t}{t^{2} + 3 + \frac{6}{x^{2}}}) d x

= \frac{2 π^{2}}{9} - 4 {[a r c t a n (\frac{x}{\sqrt{x^{2} + 2}}) a r c t a n (\sqrt{x^{2} + 2})]}_{0}^{1} + 4 \int_{0}^{1} \frac{a r c t a n (\sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 1}} d x

= 4 \int_{0}^{1} \frac{a r c t a n (\sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 1}} d x = 4 \times \frac{5 π^{2}}{96} = \frac{5 π^{2}}{24} . .

∵∴ Q = \frac{5 π^{2}}{24} \dots

\dots \dots \dots \dots \dots \dots \dots . L e p u i s s a n t \dots \dots \dots \dots \dots \dots \dots .

Commented by peter frank last updated on 18/Sep/21

good

Commented by ArielVyny last updated on 19/Sep/21

t h a n k s i r