BeMath-0-1-x-9-2-1-x-5-2-dx-

Question Number 107923 by bemath last updated on 13/Aug/20

\frac{⊚ B e M a t h ⊚}{}

\underset{0}{\int^{1}} x^{\frac{9}{2}} {(1 - x)}^{\frac{5}{2}} d x ?

Answered by 1549442205PVT last updated on 13/Aug/20

Set \frac{1}{x} - 1 = t^{2} \Rightarrow - \frac{1}{x^{2}} dx = 2 tdt \Rightarrow dx = - {2 tx}^{2} dt

t (x = 1) = 0, t (x = 0) = \infty

x = \frac{1}{\sqrt{1 + t^{2}}} \Rightarrow I = \int_{0}^{\infty} x^{\frac{9}{2}} {({xt}^{2})}^{\frac{1}{2}} {2 tx}^{2} dt

= 2 \int_{0}^{\infty} x^{7} t^{2} dt = 2 \int_{0}^{\infty} \frac{t^{2} dt}{\sqrt{{(1 + t^{2})}^{7}}}

= 2 \int_{1}^{\infty} \frac{t^{2} + 1 - 1}{{(1 + t^{2})}^{3} \sqrt{1 + t^{2}}} dt = 2 \int \frac{dt}{{(1 + t^{2})}^{2} \sqrt{1 + t^{2}}}

- 2 \int_{0}^{\infty} \frac{dt}{{(1 + t^{2})}^{3} \sqrt{1 + t^{2}}} =

Set t = \tan φ \Rightarrow dt = (1 + t^{2}) d φ

I = 2 \int_{0}^{\frac{π}{2}} \frac{d φ}{(1 + \tan^{2} φ) \sqrt{1 + \tan^{2} φ)}}

- 2 \int_{0}^{\frac{π}{2}} \frac{d φ}{{(1 + \tan^{2} φ)}^{2} \sqrt{1 + \tan^{2} φ)}}

= 2 \int_{0}^{\frac{π}{2}} \cos^{3} φ d φ - 2 \int_{0}^{\frac{π}{2}} \cos^{5} d φ

= 2 \int_{0}^{\frac{π}{2}} (1 - \sin^{2} φ) dsin φ - 2 \int_{0}^{\frac{π}{2}} {(1 - \sin^{2} φ)}^{2} dsin φ

= (2 \sin φ - \frac{{2 \sin}^{3} φ}{3}) ∣_{0}^{\frac{π}{2}} - 2 \int_{0}^{\frac{π}{2}} [1 - {2 \sin}^{2} φ + \sin^{4} φ) dsin φ

= 2 - \frac{2}{3} - 2 {[\sin φ - \frac{2}{3} \sin^{3} φ + \frac{\sin^{5} φ}{5}]}_{0}^{\frac{π}{2}}

= 2 - \frac{2}{3} - 2 (1 - \frac{2}{3} + \frac{1}{5}) = \frac{4}{15}

Answered by bobhans last updated on 13/Aug/20

\frac{∦ B ob H ans ∦}{Π}

I = - \frac{2}{7} \underset{0}{\int^{1}} x^{9 / 2} d {(1 - x)}^{7 / 2} = - \frac{2}{7} {[x^{9 / 2} {(1 - x)}^{7 / 2}]}_{0}^{1} +

\frac{9}{7} \underset{0}{\int^{1}} x^{7 / 2} {(1 - x)}^{7 / 2} dx

let J = \underset{0}{\int^{1}} x^{7 / 2} {(1 - x)}^{7 / 2} dx = \frac{9}{7} \underset{0}{\int^{1}} {(x - x^{2})}^{7 / 2} dx

J = \frac{9}{7} \underset{0}{\int^{1}} {[\frac{1}{4} - {(\frac{1}{2} - x)}^{2}]}^{7 / 2} dx

set \frac{1}{2} - x = z

J = \frac{9}{7 \times 2^{8}} \underset{- π / 2}{\int^{π / 2}} \cos^{8} z dz

Reduction formula

J_{n} = \frac{n - 1}{n} . J_{n - 2} \Rightarrow J_{8} = \frac{7}{8} . \frac{5}{6} . \frac{3}{4} . \frac{1}{2} . J_{0}

J_{8} = \frac{35}{128} . [\frac{π}{2} - (- \frac{π}{2})] = \frac{35 π}{128}

conclusion

I = \frac{9}{7 \times 2^{8}} \times \frac{35 π}{128} = \frac{45 π}{32768} .

Commented by bemath last updated on 14/Aug/20

c o r r e c t \dots . t h a n k y o u

Answered by mathmax by abdo last updated on 13/Aug/20

I = \int_{0}^{1} x^{\frac{9}{2}} {(1 - x)}^{\frac{5}{2}} dx we know B (p, q) = \int_{0}^{1} x^{p - 1} {(1 - x)}^{q - 1} dx

= \frac{Γ (p) . Γ (q)}{Γ (p + q)} here p = \frac{9}{2} + 1 and q = \frac{5}{2} + 1 \Rightarrow

I = \frac{Γ (\frac{9}{2} + 1) Γ (\frac{5}{2} + 1)}{Γ (\frac{9}{2} + \frac{5}{2} + 2)} = \frac{9}{2} . \frac{5}{2} . \frac{Γ (\frac{9}{2}) Γ (\frac{5}{2})}{Γ (9)} = \frac{45}{4.8!} Γ (\frac{9}{2}) Γ (\frac{5}{2})

Γ (\frac{9}{2}) = Γ (1 + \frac{7}{2}) = \frac{7}{2} Γ (\frac{7}{2}) = \frac{7}{2} Γ (1 + \frac{5}{2}) = \frac{7}{2} . \frac{5}{2} Γ (\frac{5}{2})

= \frac{35}{4} Γ (1 + \frac{3}{2}) = \frac{35}{4} . \frac{3}{2} Γ (\frac{3}{2}) = \frac{3.35}{8} Γ (\frac{1}{2} + 1) = \frac{3.35}{16} Γ (\frac{1}{2})

we follow the same way to find Γ (\frac{5}{2})

Commented by bemath last updated on 14/Aug/20

t y p o i t s h o u l d b e 8! i n d e n u m e r a t o r

Facebook Tweet Pin