use-beta-function-to-calculate-0-pi-sin-3-x-2-cosx-6-dx-

Question Number 100087 by mathmax by abdo last updated on 24/Jun/20

use beta function to calculate \int_{0}^{π} \sin^{3} x {(2 + cosx)}^{6} dx

Commented by bemath last updated on 25/Jun/20

\int (\cos^{2} x - 1) {(2 + \cos x)}^{6} d (\cos x)

\int (z^{2} - 1) {(2 + z)}^{6} dz

\int z^{2} {(2 + z)}^{6} dz - \int {(2 + z)}^{6} dz

{[\frac{1}{7} z^{2} {(2 + z)}^{7} - \frac{1}{28} z {(2 + z)}^{8} + \frac{1}{494} {(2 + z)}^{9} - \frac{1}{7} {(2 + z)}^{7}]}_{1}^{- 1}

[\frac{1}{7} + \frac{1}{28} + \frac{1}{494} - \frac{1}{7}] - [\frac{3^{7}}{7} - \frac{3^{8}}{28} + \frac{3^{9}}{494} - \frac{3^{7}}{7}]

= \frac{3^{8} + 1}{28} + \frac{1 - 3^{9}}{494}

Answered by 1549442205 last updated on 26/Jun/20

F = - \int_{0}^{π} (1 - \cos^{2} x) {(2 + \cos x)}^{6} dcos x

\underset{t = \cos x}{=} \int (t^{2} - 1) {(t + 2)}^{6} dt = \int [t^{2} {(t + 2)}^{6}] d \int t - \int {(t + 2)}^{6} dt = A - B

A = \int t^{2} (t^{6} + {12 t}^{5} + {60 t}^{4} + {160 t}^{3} + {240 t}^{2} + 192 t + 64) dt

= \frac{t^{9}}{9} + \frac{{12 t}^{8}}{8} + \frac{{60 t}^{7}}{7} + \frac{{160 t}^{6}}{6} + \frac{{240 t}^{5}}{5} + \frac{{192 t}^{4}}{4} + \frac{{64 t}^{3}}{3}

B = \frac{t^{7}}{7} + \frac{{12 t}^{6}}{6} + \frac{{60 t}^{5}}{5} + \frac{{160 t}^{4}}{4} + \frac{{240 t}^{3}}{3} + \frac{{192 t}^{2}}{2} + 64 t

Hence, \int_{0}^{π} \sin^{3} x {(2 + \cos x)}^{6} dx = (A - B) ∣_{1}^{- 1}

= (\frac{t^{9}}{9} + \frac{{12 t}^{8}}{8} + \frac{{59 t}^{7}}{7} + \frac{{148 t}^{6}}{6} + \frac{{180 t}^{5}}{5} + \frac{{32 t}^{4}}{4} - \frac{{176 t}^{3}}{3} - \frac{{192 t}^{2}}{2} - 64 t) ∣_{1}^{- 1}

= \frac{2053}{126} - (- \frac{17635}{126}) = \frac{19688}{126} = \frac{9844}{63} \approx 156.25

The second way :

F = \int_{0}^{π} \sin^{3} x {(2 + \cos x)}^{6} dx = \int_{0}^{π} \sin^{3} x (\cos^{6} x + {12 \cos}^{5} x + {60 \cos}^{4} x + {160 \cos}^{3} x + {240 \cos}^{2} x + 192 \cos x + 64) dx

Since, \int_{0}^{π} \sin^{2 m + 1} {xcos}^{2 n + 1} xdx =

= \int_{0}^{\frac{π}{2}} \sin^{2 m + 1} {xcos}^{2 n + 1} xdx + \int_{\frac{π}{2}}^{π} \sin^{2 m + 1} x . \cos^{2 n + 1} xdx

\underset{x = t + \frac{π}{2}}{=} \int_{0}^{\frac{π}{2}} \sin^{2 m + 1} x . \cos^{2 n + 1} xdx - \int_{0}^{\frac{π}{2}} \cos^{2 m + 1} t . \sin^{2 n + 1} tdt

\frac{1}{2} B (m + 1, n + 1) - \frac{1}{2} B (n + 1, m + 1) = 0 (due to

the property of Beta function : B (p, q) = B (q, p)

and, \int_{0}^{π} \sin^{2 m} {xcos}^{2 n + 1} xdx =

= \int_{0}^{\frac{π}{2}} \sin^{2 m} {xcos}^{2 n + 1} xdx + \int_{\frac{π}{2}}^{π} \sin^{2 m} x . \cos^{2 n + 1} xdx

== \int_{0}^{\frac{π}{2}} \sin^{2 m} {xcos}^{2 n + 1} xdx + \int_{0}^{\frac{π}{2}} \cos^{2 m} x . \sin^{2 n + 1} xdx =

\frac{1}{2} B (\frac{2 m + 1}{2}, n + 1) + \frac{1}{2} B (n + 1, \frac{2 m + 1}{2}) = B (\frac{2 m + 1}{2}, n + 1), so

F = \int_{0}^{π} \sin^{3} x (\cos^{6} x + {12 \cos}^{5} x + {60 \cos}^{4} x + {160 \cos}^{3} x + {240 \cos}^{2} x + 192 \cos x + 64) dx

= B (2, \frac{7}{2}) + 60 B (2, \frac{5}{2}) + 240 B (2, \frac{3}{2}) + 64 B (2, \frac{1}{2})

B (p, q) = \frac{Γ (p) . Γ (q)}{Γ (p + q)}, so B (2, \frac{7}{2}) = \frac{Γ (2) . Γ (\frac{7}{2})}{Γ (2 + \frac{7}{2})}

Γ (2) = 1, Γ (\frac{1}{2}) = \sqrt{π}, Γ (\frac{3}{2}) = \frac{\sqrt{π}}{2}, Γ (\frac{5}{2}) = \frac{3 \sqrt{π}}{4}, Γ (\frac{7}{2}) = \frac{15 \sqrt{π}}{8}

\Rightarrow Γ (2 + \frac{7}{2}) = Γ (5 + \frac{1}{2}) = \frac{3.5.7.9}{2^{5}} \sqrt{π} = \frac{945 \sqrt{π}}{32}

\Rightarrow B (2, \frac{7}{2}) = \frac{1 . \frac{15 \sqrt{π}}{8}}{\frac{945 \sqrt{π}}{32}} = \frac{12}{189} = \frac{4}{63}

B (2, \frac{5}{2}) = \frac{Γ (2) . Γ (\frac{5}{2})}{Γ (2 + \frac{5}{2})} = \frac{1 . \frac{3 \sqrt{π}}{4}}{Γ (4 + \frac{1}{2})} = \frac{\frac{3 \sqrt{π}}{4}}{\frac{3.5.7 \sqrt{π}}{16}} = \frac{4}{35}

B (2, \frac{3}{2}) = \frac{Γ (2) . Γ (\frac{3}{2})}{Γ (2 + \frac{3}{2})} = \frac{\frac{\sqrt{π}}{2}}{Γ (3 + \frac{1}{2})} = \frac{\frac{\sqrt{π}}{2}}{\frac{3.5 \sqrt{π}}{8}} = \frac{4}{15}

B (2, \frac{1}{2}) = \frac{Γ (2) . Γ (\frac{1}{2})}{Γ (2 + \frac{1}{2})} = \frac{\sqrt{π}}{\frac{3 \sqrt{π}}{4}} = \frac{4}{3} .

Therefore, F = \frac{4}{63} + 60 . \frac{4}{35} + 240 . \frac{4}{15} + 64 . \frac{4}{3}

= \frac{9844}{63}

Commented by mathmax by abdo last updated on 25/Jun/20

thank you sir for this hardwork . .