∫_0 ^(π/2) arcos(((cosx)/(1+2cosx)))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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

	$Q = \int_{0}^{\frac{π}{2}} a r c c o s (\frac{c o s x}{1 + 2 c o s x}) d x = 2 \int_{0}^{\frac{π}{2}} a r c c o s (\sqrt{\frac{1 + \frac{c o s x}{1 + 2 c o s x}}{2}}) d x$ $= 2 \int_{0}^{\frac{π}{2}} a r c t a n (\sqrt{\frac{1 + c o s x}{1 + 3 c o s x}}) d x = 4 \int_{0}^{\frac{π}{4}} a r c t a n (\sqrt{\frac{1 + c o s 2 u}{1 + 3 c o s 2 u}}) d u$ $D i v e r :^{″} 1 + c o s 2 u = 2 c o s u e t c o s 2 u = 1 - {s i n}^{2} u {(T r i v i a l)}^{″}$ $\Rightarrow Q = 4 \int_{0}^{\frac{π}{4}} a r c t a n (\sqrt{\frac{{c o s}^{2} u}{1 - {s i n}^{2} u}}) d u = 4 \int_{0}^{\frac{π}{4}} a r c t a n (\frac{c o s u}{\sqrt{1 - 2 s i n u}}) d u$ $= 4 \int_{0}^{\frac{π}{4}} \int_{0}^{1} \frac{\sqrt{2 - 3 {s i n}^{2} u} c o s u}{(x^{2} + 2) - {s i n}^{2} u (x^{2} + 3)} d x d u$ $\sqrt{3} s i n u = \sqrt{2} s i n θ \to c o s u d u = \frac{\sqrt{2}}{\sqrt{3}} c o s θ d θ$ $\Rightarrow Q = 4 \int_{0}^{\frac{π}{3}} \int_{0}^{1} \frac{2 \sqrt{3} {c o s}^{2} θ}{3 x^{2} + 6 - 2 (1 - {c o s}^{2} θ) (x^{2} + 3)} d x d θ$ $\overset{t = t a n u}{=} 8 \sqrt{3} \int_{0}^{\sqrt{3}} \int_{0}^{1} \frac{1}{(1 + t^{2}) (t^{2} x^{2} + 3 x^{2} + 6)} d x d t$ $= 8 \sqrt{3} \int_{0}^{\sqrt{3}} \int_{0}^{1} (\frac{- \frac{1}{2 x^{2} + 6}}{1 + t^{2}} + \frac{\frac{x^{2}}{2 x^{2} + 6}}{t^{2} x^{2} + 3 x^{2} + 6}) d x d t = 8 \sqrt{3} \int_{0}^{1} \frac{1}{2 x^{2} + 6} (\int_{0}^{\sqrt{3}} \frac{d t}{1 + t^{2}} - \int_{0}^{\sqrt{3}} \frac{d t}{t^{2} + 3 + \frac{6}{x^{2}}}) d x$ $= \frac{2 π^{2}}{9} - 4 {[a r c t a n (\frac{x}{\sqrt{x^{2} + 2}}) a r c t a n (\sqrt{x^{2} + 2})]}_{0}^{1} + 4 \int_{0}^{1} \frac{a r c t a n (\sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 1}} d x$ $= 4 \int_{0}^{1} \frac{a r c t a n (\sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 1}} d x = 4 \times \frac{5 π^{2}}{96} = \frac{5 π^{2}}{24} . .$ $∵∴ Q = \frac{5 π^{2}}{24} . . .$ $. . . . . . . . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . . . . . . . . .$
	Commented by peter frank last updated on 18/Sep/21

	$good$
	Commented by ArielVyny last updated on 19/Sep/21

	$t h a n k s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum