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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

case one a=0 ∫_0 ^(2a) x^n (√(2ax−x^2 ))dx=0 other case x=2ay ∫_0 ^(2a) x^n (√(2ax−x^2 ))dx=∫_0 ^1 (2ay)^n (√(4a^2 (y−y^2 ) )) 2ady,a∈R =(2a)^(n+1) ∫_0 ^1 y^n ∣2a∣(√(y−y^2 ))dy =Sign(a)(2a)^(n+2) ∫_0 ^1 y^n (√(y−y^2 ))dy=sign(a)(2a)^(n+2) ∫_0 ^1 y^n .(√y).(√(1−y))dy ∫_0 ^1 y^n .(√y).(√(1−y))dy=∫_0 ^1 y^(n+(1/2)) (1−y)^(1/2) dy=β(n+(3/2),(3/2)) =((Γ(n+(3/2))Γ((3/2)))/(Γ(n+3)))=((Γ(n+(3/2)).(√π))/(2(n+2)!)) we get∫_0 ^(2a) x^n (√(2ax−x^2 ))dx=sign(a)(2a)^(n+2) ∫_0 ^1 y^n (√(y−y^2 ))dy =sign(a)(2a)^(n+2) .((Γ(n+(3/2))(√π))/(2.(n+2)!))

$${case}\:{one}\:{a}=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{a}} {x}^{{n}} \sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$${other}\:{case}\:{x}=\mathrm{2}{ay} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}{a}} {x}^{{n}} \sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{2}{ay}\right)^{{n}} \sqrt{\mathrm{4}{a}^{\mathrm{2}} \left({y}−{y}^{\mathrm{2}} \right)\:}\:\:\mathrm{2}{ady},{a}\in\mathbb{R} \\ $$$$=\left(\mathrm{2}{a}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} \mid\mathrm{2}{a}\mid\sqrt{{y}−{y}^{\mathrm{2}} }{dy} \\ $$$$={Sign}\left({a}\right)\left(\mathrm{2}{a}\right)^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} \sqrt{{y}−{y}^{\mathrm{2}} }{dy}={sign}\left({a}\right)\left(\mathrm{2}{a}\right)^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} .\sqrt{{y}}.\sqrt{\mathrm{1}−{y}}{dy} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} .\sqrt{{y}}.\sqrt{\mathrm{1}−{y}}{dy}=\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{y}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dy}=\beta\left({n}+\frac{\mathrm{3}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left({n}+\mathrm{3}\right)}=\frac{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right).\sqrt{\pi}}{\mathrm{2}\left({n}+\mathrm{2}\right)!} \\ $$$${we}\:{get}\int_{\mathrm{0}} ^{\mathrm{2}{a}} {x}^{{n}} \sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }{dx}={sign}\left({a}\right)\left(\mathrm{2}{a}\right)^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {y}^{{n}} \sqrt{{y}−{y}^{\mathrm{2}} }{dy} \\ $$$$={sign}\left({a}\right)\left(\mathrm{2}{a}\right)^{{n}+\mathrm{2}} .\frac{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{2}.\left({n}+\mathrm{2}\right)!} \\ $$

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

$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{2a}} \:\mathrm{x}^{\mathrm{n}} \sqrt{\mathrm{2ax}−\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:\Rightarrow\:\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{2a}} \:\mathrm{x}^{\mathrm{n}} \sqrt{\mathrm{x}}\sqrt{\mathrm{2a}−\mathrm{x}}\mathrm{dx} \\ $$$$=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\:\:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2a}}} \:\:\mathrm{t}^{\mathrm{2n}} \mathrm{t}\sqrt{\mathrm{2a}−\mathrm{t}^{\mathrm{2}} }\mathrm{2t}\:\mathrm{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2a}}} \mathrm{t}^{\mathrm{2n}+\mathrm{2}} \sqrt{\left(\sqrt{\mathrm{2a}}\right)^{\mathrm{2}} −\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=_{\mathrm{t}=\sqrt{\mathrm{2a}}\mathrm{sin}\theta} \:\:\:\:\mathrm{2}\:\:\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \:\mathrm{sin}^{\mathrm{2n}+\mathrm{2}} \theta\sqrt{\mathrm{2a}}\mathrm{cost}\:\sqrt{\mathrm{2a}}\mathrm{cost}\:\mathrm{dt} \\ $$$$=\mathrm{4a}\:\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \:\:\:\mathrm{sin}^{\mathrm{2n}+\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\mathrm{4a}\:\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \mathrm{sin}^{\mathrm{2n}+\mathrm{2}} \mathrm{t}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}\right)\mathrm{dt} \\ $$$$=\mathrm{4a}\:\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \:\mathrm{sin}^{\mathrm{2n}+\mathrm{2}} \mathrm{t}\:−\mathrm{4a}\:\int_{\mathrm{0}} ^{\mathrm{arcsinA}} \:\:\mathrm{sin}^{\mathrm{2n}+\mathrm{4}} \mathrm{tdt} \\ $$$$\Rightarrow\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\mathrm{I}_{\mathrm{k}} =\mathrm{4a}\:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \:\left(\mathrm{v}_{\mathrm{k}} −\mathrm{v}_{\mathrm{k}+\mathrm{1}} \right)\:\mathrm{with}\:\mathrm{v}_{\mathrm{k}} =\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \:\mathrm{sin}^{\mathrm{2k}+\mathrm{2}} \:\mathrm{tdt} \\ $$$$=\mathrm{4a}\left(\mathrm{v}_{\mathrm{0}} −\mathrm{v}_{\mathrm{1}} +\mathrm{v}_{\mathrm{1}} −\mathrm{v}_{\mathrm{2}} +....+\mathrm{v}_{\mathrm{n}} −\mathrm{v}_{\mathrm{n}+\mathrm{1}} \right) \\ $$$$=\mathrm{4a}\:\left(\mathrm{v}_{\mathrm{0}} −\mathrm{v}_{\mathrm{n}+\mathrm{1}} \right)\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{v}_{\mathrm{0}} =\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2t}\right)\right)\mathrm{dt}\:=\frac{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)}{\mathrm{2}}\:+\left[\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2t}\right)\right]_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \\ $$$$=\frac{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sint}\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\right]_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \\ $$$$=\frac{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\sqrt{\mathrm{2a}}\sqrt{\mathrm{1}−\mathrm{2a}}\right\} \\ $$$$\mathrm{v}_{\mathrm{n}+\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)} \mathrm{sin}^{\mathrm{2n}+\mathrm{4}} \mathrm{t}\:\mathrm{dt}\:\:\left(\mathrm{wallis}\:\mathrm{integral}\:\mathrm{on}\left[\mathrm{0},\mathrm{arcsin}\left(\sqrt{\mathrm{2a}}\right)\right)\right. \\ $$$$\mathrm{also}\:\mathrm{we}\:\mathrm{can}\:\mathrm{determine}\:\mathrm{a}\:\mathrm{recurrence}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{I}_{\mathrm{n}} \\ $$

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