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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
	$I_n =∫_0 ^(2a) x^n (√(2ax−x^2 )) dx ⇒ I_n =∫_0 ^(2a) x^n (√x)(√(2a−x))dx =_((√x)=t) ∫_0 ^(√(2a)) t^(2n) t(√(2a−t^2 ))2t dt =2 ∫_0 ^(√(2a)) t^(2n+2) (√(((√(2a)))^2 −t^2 ))dt =_(t=(√(2a))sinθ) 2 ∫_0 ^(arcsin((√(2a)))) sin^(2n+2) θ(√(2a))cost (√(2a))cost dt =4a ∫_0 ^(arcsin((√(2a)))) sin^(2n+2) cos^2 t dt =4a ∫_0 ^(arcsin((√(2a)))) sin^(2n+2) t(1−sin^2 t)dt =4a ∫_0 ^(arcsin((√(2a)))) sin^(2n+2) t −4a ∫_0 ^(arcsinA) sin^(2n+4) tdt ⇒Σ_(k=0) ^n I_k =4a Σ_(k=0) ^n (v_k −v_(k+1) ) with v_k =∫_0 ^(arcsin((√(2a)))) sin^(2k+2) tdt =4a(v_0 −v_1 +v_1 −v_2 +....+v_n −v_(n+1) ) =4a (v_0 −v_(n+1) ) we have v_0 =∫_0 ^(arcsin((√(2a)))) sin^2 t dt =(1/2)∫_0 ^(arcsin((√(2a)))) (1+cos(2t))dt =((arcsin((√(2a))))/2) +[(1/4)sin(2t)]_0 ^(arcsin((√(2a)))) =((arcsin((√(2a))))/2) +(1/2)[sint (√(1−sin^2 t))]_0 ^(arcsin((√(2a)))) =((arcsin((√(2a))))/2) +(1/2){(√(2a))(√(1−2a))} v_(n+1) =∫_0 ^(arcsin((√(2a)))) sin^(2n+4) t dt (wallis integral on[0,arcsin((√(2a)))) also we can determine a recurrence relation between I_n$
	$I_{n} = \int_{0}^{2 a} x^{n} \sqrt{2 ax - x^{2}} dx \Rightarrow I_{n} = \int_{0}^{2 a} x^{n} \sqrt{x} \sqrt{2 a - x} dx$ $=_{\sqrt{x} = t} \int_{0}^{\sqrt{2 a}} t^{2 n} t \sqrt{2 a - t^{2}} 2 t dt = 2 \int_{0}^{\sqrt{2 a}} t^{2 n + 2} \sqrt{{(\sqrt{2 a})}^{2} - t^{2}} dt$ $=_{t = \sqrt{2 a} \sin θ} 2 \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} θ \sqrt{2 a} cost \sqrt{2 a} cost dt$ $= 4 a \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} \cos^{2} t dt$ $= 4 a \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} t (1 - \sin^{2} t) dt$ $= 4 a \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 2} t - 4 a \int_{0}^{arcsinA} \sin^{2 n + 4} tdt$ $\Rightarrow \sum_{k = 0}^{n} I_{k} = 4 a \sum_{k = 0}^{n} (v_{k} - v_{k + 1}) with v_{k} = \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 k + 2} tdt$ $= 4 a (v_{0} - v_{1} + v_{1} - v_{2} + . . . . + v_{n} - v_{n + 1})$ $= 4 a (v_{0} - v_{n + 1}) we have v_{0} = \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2} t dt$ $= \frac{1}{2} \int_{0}^{\arcsin (\sqrt{2 a})} (1 + \cos (2 t)) dt = \frac{\arcsin (\sqrt{2 a})}{2} + {[\frac{1}{4} \sin (2 t)]}_{0}^{\arcsin (\sqrt{2 a})}$ $= \frac{\arcsin (\sqrt{2 a})}{2} + \frac{1}{2} {[sint \sqrt{1 - \sin^{2} t}]}_{0}^{\arcsin (\sqrt{2 a})}$ $= \frac{\arcsin (\sqrt{2 a})}{2} + \frac{1}{2} {\sqrt{2 a} \sqrt{1 - 2 a}}$ $v_{n + 1} = \int_{0}^{\arcsin (\sqrt{2 a})} \sin^{2 n + 4} t dt (wallis integral on [0, \arcsin (\sqrt{2 a}))$ $also we can determine a recurrence relation between I_{n}$
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