Use-gamma-function-to-prove-i-0-pi-4-sin-4-x-2x-dx-3-4-192-ii-0-pi-6-cos-4-3-sin-2-6-5pi-192-

Question Number 82560 by niroj last updated on 22/Feb/20

U se gamma function to prove

(i) \int_{0}^{\frac{π}{4}} \sin^{4} x 2 x dx = \frac{3 π - 4}{192} .

(ii) \int_{0}^{\frac{π}{6}} \cos^{4} 3 θ \sin^{2} 6 θ = \frac{5 π}{192} .

Answered by M±th+et£s last updated on 22/Feb/20

N . B : Γ (z + \frac{1}{2}) = \frac{(2 z - 1)!!}{2^{z}} \sqrt{π} β = \frac{Γ (x) Γ (y)}{Γ (x + y)} = \int_{0}^{\frac{π}{2}} {c o s}^{2 x - 1} θ {s i n}^{2 x - 1} θ d θ

I = \int_{0}^{\frac{π}{6}} {c o s}^{4} 3 θ {s i n}^{2} 6 θ d θ \overset{θ \to 3 θ}{=} \frac{1}{3} \int_{0}^{\frac{π}{2}} {c o s}^{4} θ {s i n}^{2} 2 θ d θ

= \frac{4}{3} \int_{0}^{\frac{π}{2}} {c o s}^{6} θ {s i n}^{2} θ d θ = \frac{2}{3} β (\frac{7}{2}, \frac{3}{2})

I = \frac{2}{3} \frac{Γ (\frac{7}{2}, \frac{3}{2})}{Γ (5)} = \frac{2}{3} \frac{Γ (3 + \frac{1}{2}) Γ (1 + \frac{1}{2})}{Γ (5)}

= \frac{2}{3} \frac{(\frac{5!!}{2^{3}} \sqrt{π}) (\frac{1!!}{2^{1}} \sqrt{π})}{4!} = \frac{2 π}{3} \frac{(\frac{15}{8}) (\frac{1}{2})}{24}

I = \frac{5 π}{192}