calculate-0-pi-4-1-3-sin-x-cos-x-n-dx-1-n-

Question Number 191342 by mnjuly1970 last updated on 23/Apr/23

c a l c u l a t e \dots

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

\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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