Question-116710

Question Number 116710 by ravisoni505 last updated on 06/Oct/20

Answered by maths mind last updated on 06/Oct/20

c a s e o n e a = 0

\int_{0}^{2 a} x^{n} \sqrt{2 a x - x^{2}} d x = 0

o t h e r c a s e x = 2 a y

\int_{0}^{2 a} x^{n} \sqrt{2 a x - x^{2}} d x = \int_{0}^{1} {(2 a y)}^{n} \sqrt{4 a^{2} (y - y^{2})} 2 a d y, a \in R

= {(2 a)}^{n + 1} \int_{0}^{1} y^{n} ∣ 2 a ∣ \sqrt{y - y^{2}} d y

= S i g n (a) {(2 a)}^{n + 2} \int_{0}^{1} y^{n} \sqrt{y - y^{2}} d y = s i g n (a) {(2 a)}^{n + 2} \int_{0}^{1} y^{n} . \sqrt{y} . \sqrt{1 - y} d y

\int_{0}^{1} y^{n} . \sqrt{y} . \sqrt{1 - y} d y = \int_{0}^{1} y^{n + \frac{1}{2}} {(1 - y)}^{\frac{1}{2}} d y = β (n + \frac{3}{2}, \frac{3}{2})

= \frac{Γ (n + \frac{3}{2}) Γ (\frac{3}{2})}{Γ (n + 3)} = \frac{Γ (n + \frac{3}{2}) . \sqrt{π}}{2 (n + 2)!}

w e g e t \int_{0}^{2 a} x^{n} \sqrt{2 a x - x^{2}} d x = s i g n (a) {(2 a)}^{n + 2} \int_{0}^{1} y^{n} \sqrt{y - y^{2}} d y

= s i g n (a) {(2 a)}^{n + 2} . \frac{Γ (n + \frac{3}{2}) \sqrt{π}}{2 . (n + 2)!}

Answered by mathmax by abdo last updated on 06/Oct/20

I_{n} = \int_{0}^{2 a} x^{n} \sqrt{2 ax - x^{2}} dx \Rightarrow I_{n} = \int_{0}^{2 a} x^{n} \sqrt{x} \sqrt{2 a - x} dx

=_{\sqrt{x} = t} \int_{0}^{\sqrt{2 a}} t^{2 n} t \sqrt{2 a - t^{2}} 2 t dt = 2 \int_{0}^{\sqrt{2 a}} t^{2 n + 2} \sqrt{{(\sqrt{2 a})}^{2} - t^{2}} dt

=_{t = \sqrt{2 a} \sin θ} 2 \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} θ \sqrt{2 a} cost \sqrt{2 a} cost dt

= 4 a \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} \cos^{2} t dt

= 4 a \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} t (1 - \sin^{2} t) dt

= 4 a \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} t - 4 a \int_{0}^{arcsinA} \sin^{2 n + 4} tdt

\Rightarrow \sum_{k = 0}^{n} I_{k} = 4 a \sum_{k = 0}^{n} (v_{k} - v_{k + 1}) with v_{k} = \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 k + 2} tdt

= 4 a (v_{0} - v_{1} + v_{1} - v_{2} + \dots . + v_{n} - v_{n + 1})

= 4 a (v_{0} - v_{n + 1}) we have v_{0} = \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2} t dt

= \frac{1}{2} \int_{0}^{\arcsin (\sqrt{2 a})} (1 + \cos (2 t)) dt = \frac{\arcsin (\sqrt{2 a})}{2} + {[\frac{1}{4} \sin (2 t)]}_{0}^{\arcsin (\sqrt{2 a})}

= \frac{\arcsin (\sqrt{2 a})}{2} + \frac{1}{2} {[sint \sqrt{1 - \sin^{2} t}]}_{0}^{\arcsin (\sqrt{2 a})}

= \frac{\arcsin (\sqrt{2 a})}{2} + \frac{1}{2} {\sqrt{2 a} \sqrt{1 - 2 a}}

v_{n + 1} = \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 4} t dt (wallis integral on [0, \arcsin (\sqrt{2 a}))

also we can determine a recurrence relation between I_{n}