by-use-Gamma-function-prove-1-0-pi-8-cos-3-4xdx-1-6-2-0-pi-sin-6-x-2-cos-8-x-2-dx-5pi-2-11-

Question Number 139371 by mohammad17 last updated on 26/Apr/21

b y u s e G a m m a f u n c t i o n p r o v e

(1) \int_{0}^{\frac{π}{8}} {c o s}^{3} 4 x d x = \frac{1}{6}

(2) \int_{0}^{π} {s i n}^{6} (\frac{x}{2}) {c o s}^{8} (\frac{x}{2}) d x = \frac{5 π}{2^{11}}

Answered by Dwaipayan Shikari last updated on 26/Apr/21

\frac{x}{2} = u

\Rightarrow 2 \int_{0}^{\frac{π}{2}} {s i n}^{6} (u) {c o s}^{8} (u) d u = \frac{Γ (\frac{7}{2}) Γ (\frac{9}{2})}{Γ (8)} = \frac{\frac{5}{2} . \frac{3}{2} . \frac{3}{2} . \frac{5}{2} . \frac{7}{2} . \frac{1}{2} . \frac{1}{2} π}{7!}

= \frac{π}{2^{7}} . \frac{5.5.3.3}{6!} = \frac{π}{2^{8}} . \frac{3.5}{4!} = \frac{5 π}{2^{11}}

Commented by mohammad17 last updated on 26/Apr/21

t h a n k y o u s i r a n d n u m b e r (1) c a n y o u s o l v e

Answered by Dwaipayan Shikari last updated on 26/Apr/21

4 x = u

\Rightarrow \frac{1}{4} \int_{0}^{\frac{π}{2}} {c o s}^{3} (u) d u = \frac{1}{4} . \frac{Γ (2) Γ (\frac{1}{2})}{2 Γ (\frac{5}{2})} = \frac{1}{4} . \frac{\sqrt{π}}{\frac{3 \sqrt{π}}{4}} . \frac{1}{2} = \frac{1}{6}

Commented by mohammad17 last updated on 26/Apr/21

s i r c a n y o u c l e a r t h i s s t e b h o w g e t

Γ (2) a n d Γ (\frac{1}{2}) f r o m {c o s}^{3} (u)

Commented by mohammad17 last updated on 26/Apr/21

t h a n k y o u s i r

Commented by Dwaipayan Shikari last updated on 26/Apr/21

\int_{0}^{\frac{π}{2}} {c o s}^{2 n - 1} x {s i n}^{2 α - 1} x d x = \frac{Γ (n) Γ (α)}{2 Γ (n + α)}

\int_{0}^{\frac{π}{2}} {c o s}^{3} x d x = \int_{0}^{\frac{π}{2}} {c o s}^{2.2 - 1} x {s i n}^{2 (\frac{1}{2}) - 1} x d x = \frac{Γ (2) Γ (\frac{1}{2})}{2 Γ (2 + \frac{1}{2})}

Answered by Ar Brandon last updated on 26/Apr/21

I = \int_{0}^{\frac{π}{8}} \cos^{3} 4 xdx, u = 4 x

= \frac{1}{4} \int_{0}^{\frac{π}{2}} \cos^{3} udu = \frac{1}{8} β (2, \frac{1}{2})

= \frac{1}{8} \cdot \frac{Γ (2) Γ (\frac{1}{2})}{Γ (\frac{5}{2})} = \frac{1}{8} \cdot \frac{\sqrt{π}}{\frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{π}} = \frac{1}{6}

J = \int_{0}^{π} \sin^{6} (\frac{x}{2}) \cos^{8} (\frac{x}{2}) dx, x = 2 v

= 2 \int_{0}^{\frac{π}{2}} \sin^{6} {vcos}^{8} vdv = β (\frac{7}{2}, \frac{9}{2})

= \frac{Γ (\frac{7}{2}) Γ (\frac{9}{2})}{Γ (8)} = \frac{1}{7!} \cdot {(\frac{5}{2} \cdot \frac{3}{2} \cdot \frac{\sqrt{π}}{2})}^{2} \cdot \frac{7}{2}

= \frac{225 π}{720 \times 2^{7}} = \frac{5 π}{2048} = \frac{5 π}{2^{11}}