calculate... Ω ={ ∫_0 ^( (π/4)) ( 1 + (√3) sin(x) + cos(x) )^( n) dx}^(1/n) = ?


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Question Number 191342 by mnjuly1970 last updated on 23/Apr/23
$calculate... Ω ={ ∫_0 ^( (π/4)) ( 1 + (√3) sin(x) + cos(x) )^( n) dx}^(1/n) = ?$
$c a l c u l a t e . . .$ $Ω = {\int_{0}^{\frac{π}{4}} {(1 + \sqrt{3} s i n (x) + c o s (x))}^{n} d x}^{\frac{1}{n}} = ?$
	Answered by witcher3 last updated on 24/Apr/23
	$∫(1+(√3)sin(x)+cos(x))^n dx (√3)sin(x)+cos(x)=2sin(x+(π/6)),y=x+(π/6)=g(x) ⇔∫(1+2sin(y))^n dy= let w=sin(y) ⇔∫(((1+2w)^n )/( (√(1−w^2 ))))dw,t=1+2w⇒w=((t−1)/2) =∫(t^n /( (√((1−(((t−1)/2))^2 )))),(dt/2) =(1/2)∫t^n (((3−t)/2))^(−(1/2)) (((t+1)/2))^(−(1/2)) dt let F(s)=(1/2)∫_0 ^s t^n (((3−t)/2))^(−(1/2)) (((t+1)/2))^(−(1/2)) dt t=sz⇒dt=sdz f(s)=(1/( (√3)))s^(n+1) ∫_0 ^1 z^n (1−((sz)/3))^(−(1/2)) (1+sz)^(−(1/2)) dz We have F_1 (a;b,c;d;x;y).((Γ(a)Γ(d−a))/(Γ(d)))=∫_0 ^1 u^(a−1) (1−u)^(d−a−1) (1−ux)^(−b) (1−uy)^(−c) du Hypergeomtric Functiom of Two variable f(s)=(s^(n+1) /( (√3))).((Γ(n+1)Γ(1))/(Γ(n+2))).F_1 (n+1;(1/2),(1/2);n+2;(s/3);−s) g(x)=(((1+(√3)sin(x)+cos(x))^(n+1) )/((n+1)(√3)))F_1 (n+1;(1/2),(1/2);n+2;−2sin(x+(π/6))−1;(1/3)(2sin(x+(π/6))+1))+c ∫_0 ^(π/4) (1+(√3)sin(x)+cos(x))^n dx=g((π/4))−g(0) long expression =(1/((n+1)(√3))){((√(2+(√3)))+1)^(n+1) F_1 (n+1;(1/2),(1/2);n+2;−(1+(√((√3)+2)));((1+(√((√3)+2)))/3)) −2^(n+1) F_1 (n+1;(1/2),(1/2);n+2;−2,(2/3))}$
	$\int {(1 + \sqrt{3} \sin (x) + \cos (x))}^{n} dx$ $\sqrt{3} \sin (x) + \cos (x) = 2 \sin (x + \frac{π}{6}), y = x + \frac{π}{6} = g (x)$ $\Leftrightarrow \int {(1 + 2 \sin (y))}^{n} dy =$ $let w = \sin (y)$ $\Leftrightarrow \int \frac{{(1 + 2 w)}^{n}}{\sqrt{1 - w^{2}}} dw, t = 1 + 2 w \Rightarrow w = \frac{t - 1}{2}$ $= \int \frac{t^{n}}{\sqrt{(1 - {(\frac{t - 1}{2})}^{2}}}, \frac{dt}{2}$ $= \frac{1}{2} \int t^{n} {(\frac{3 - t}{2})}^{- \frac{1}{2}} {(\frac{t + 1}{2})}^{- \frac{1}{2}} dt$ $let F (s) = \frac{1}{2} \int_{0}^{s} t^{n} {(\frac{3 - t}{2})}^{- \frac{1}{2}} {(\frac{t + 1}{2})}^{- \frac{1}{2}} dt$ $t = sz \Rightarrow dt = sdz$ $f (s) = \frac{1}{\sqrt{3}} s^{n + 1} \int_{0}^{1} z^{n} {(1 - \frac{sz}{3})}^{- \frac{1}{2}} {(1 + sz)}^{- \frac{1}{2}} dz$ $We have$ $F_{1} (a; b, c; d; x; y) . \frac{Γ (a) Γ (d - a)}{Γ (d)} = \int_{0}^{1} u^{a - 1} {(1 - u)}^{d - a - 1} {(1 - ux)}^{- b} {(1 - uy)}^{- c} du$ $Hypergeomtric Functiom of Two variable$ $f (s) = \frac{s^{n + 1}}{\sqrt{3}} . \frac{Γ (n + 1) Γ (1)}{Γ (n + 2)} . F_{1} (n + 1; \frac{1}{2}, \frac{1}{2}; n + 2; \frac{s}{3}; - s)$ $g (x) = \frac{{(1 + \sqrt{3} \sin (x) + \cos (x))}^{n + 1}}{(n + 1) \sqrt{3}} F_{1} (n + 1; \frac{1}{2}, \frac{1}{2}; n + 2; - 2 \sin (x + \frac{π}{6}) - 1; \frac{1}{3} (2 \sin (x + \frac{π}{6}) + 1)) + c$ $\int_{0}^{\frac{π}{4}} {(1 + \sqrt{3} \sin (x) + \cos (x))}^{n} dx = g (\frac{π}{4}) - g (0)$ $long expression$ $= \frac{1}{(n + 1) \sqrt{3}} {{(\sqrt{2 + \sqrt{3}} + 1)}^{n + 1} F_{1} (n + 1; \frac{1}{2}, \frac{1}{2}; n + 2; - (1 + \sqrt{\sqrt{3} + 2}); \frac{1 + \sqrt{\sqrt{3} + 2}}{3})$ $- 2^{n + 1} F_{1} (n + 1; \frac{1}{2}, \frac{1}{2}; n + 2; - 2, \frac{2}{3})}$
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