Question Number 61229 by maxmathsup by imad last updated on 30/May/19
$${let}\:{f}_{{n}} \left({a}\right)\:=\int_{\mathrm{0}} ^{{a}} \:{x}^{{n}} \sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{let}\:{g}_{{n}} \left({a}\right)\:={f}^{'} \left({a}\right)\:\:\:{give}\:{g}_{{n}} \left({a}\right)\:{at}\:{form}\:{of}\:{integral}\:{and}\:{give}\:{its} \\ $$$${value}\: \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:{x}^{\mathrm{3}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}\:\:{and}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} {x}^{\mathrm{4}} \sqrt{\mathrm{3}−{x}^{\mathrm{2}} }{dx}\: \\ $$
Answered by perlman last updated on 30/May/19
![1) put x=asin(t) fn(a)=a^(n+1) ∫_0 ^(π/2) sin^n (t)(√((a^2 −a^2 sin^2 (t) )) cos(t)dt= =a^(n+2) ∫_0 ^(π/2) sin^n (t)cos^2 (t)dt=a∫sin^n (t)dt−a^(n+2) ∫sin^(n+2) (t)dt let I_n =∫_0 ^(π/2) sin^n (t)dt=∫sin(t)sin^(n−1) (t)dt=[−cos(t)sin^(n−1) (t)]+(n−1)∫cos^2 (t)sin^(n−2) (t)dt (n−1)∫_0 ^(π/2) (1−sin^2 (t))sin^((n−2)) (t)dt=(n−1)I_(n−2) −(n−1)I_n =I_n I_n =((n−1)/n)I_(n−2) I_0 =(π/2) I_1 =1 I_(2n) =((2n−1)/(2n))I_(2(n−1)) I_(2n) =((2n−1)/(2n)).((2(n−1)−1)/(2(n−1)))......((2−1)/2)I_0 =(((2n−1)(2n−3)....(1))/(2n.2(n−1)....2(1)))I_0 =((2n(2n−1)(2n−2).......1)/([2^n n!]^2 ))I_0 =(((2n)!)/(2^(2n) (n!)^2 ))(π/2) I_(2n+1) =((2n)/(2n+1))I_(2n−1) =((2n)/(2n+1)).((2n−2)/(2n−1))......(2/3)I_1 =(((2^n n!)^2 .2)/((2n+1)!))=((2^(2n+1) (n!)^2 )/((2n+1)!)) fn(a)=a^(n+2) (I_n −I_(n+2) ) gn(a)=(d/da)∫_0 ^a x^n (√((a^2 −x^2 )))dx=(d/da)∫_0 ^1 a^n t^n a^2 (√((1−t^2 )))dt=∫_0 ^1 (d/da)(a^(n+2) t^n (√((1−t^2 )) dt)= (n+2)∫_0 ^1 a^(n+1) t^n (√((1−t^2 )) dt ∫_0 ^2 x^3 (√((4−x^2 )))dx n=3 a=2 =2^5 (I1−I3)=2^5 (1−(2/3))=((32)/3)](https://www.tinkutara.com/question/Q61248.png)
$$\left.\mathrm{1}\right)\:{put}\:{x}={asin}\left({t}\right) \\ $$$${fn}\left({a}\right)={a}^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({t}\right)\sqrt{\left({a}^{\mathrm{2}} −{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)\:\right.}\:{cos}\left({t}\right){dt}= \\ $$$$={a}^{{n}+\mathrm{2}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({t}\right){cos}^{\mathrm{2}} \left({t}\right){dt}={a}\int{sin}^{{n}} \left({t}\right){dt}−{a}^{{n}+\mathrm{2}} \int{sin}^{{n}+\mathrm{2}} \left({t}\right){dt} \\ $$$${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} \left({t}\right){dt}=\int{sin}\left({t}\right){sin}^{{n}−\mathrm{1}} \left({t}\right){dt}=\left[−{cos}\left({t}\right){sin}^{{n}−\mathrm{1}} \left({t}\right)\right]+\left({n}−\mathrm{1}\right)\int{cos}^{\mathrm{2}} \left({t}\right){sin}^{{n}−\mathrm{2}} \left({t}\right){dt} \\ $$$$\left({n}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \left({t}\right)\right){sin}^{\left({n}−\mathrm{2}\right)} \left({t}\right){dt}=\left({n}−\mathrm{1}\right){I}_{{n}−\mathrm{2}} −\left({n}−\mathrm{1}\right){I}_{{n}} ={I}_{{n}} \\ $$$${I}_{{n}} =\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2}} \\ $$$${I}_{\mathrm{0}} =\frac{\pi}{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\mathrm{1} \\ $$$${I}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}{I}_{\mathrm{2}\left({n}−\mathrm{1}\right)} \\ $$$${I}_{\mathrm{2}{n}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}.\frac{\mathrm{2}\left({n}−\mathrm{1}\right)−\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}……\frac{\mathrm{2}−\mathrm{1}}{\mathrm{2}}{I}_{\mathrm{0}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)….\left(\mathrm{1}\right)}{\mathrm{2}{n}.\mathrm{2}\left({n}−\mathrm{1}\right)….\mathrm{2}\left(\mathrm{1}\right)}{I}_{\mathrm{0}} \\ $$$$=\frac{\mathrm{2}{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)…….\mathrm{1}}{\left[\mathrm{2}^{{n}} {n}!\right]^{\mathrm{2}} }{I}_{\mathrm{0}} =\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\frac{\pi}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}{n}+\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}{I}_{\mathrm{2}{n}−\mathrm{1}} =\frac{\mathrm{2}{n}}{\mathrm{2}{n}+\mathrm{1}}.\frac{\mathrm{2}{n}−\mathrm{2}}{\mathrm{2}{n}−\mathrm{1}}……\frac{\mathrm{2}}{\mathrm{3}}{I}_{\mathrm{1}} =\frac{\left(\mathrm{2}^{{n}} {n}!\right)^{\mathrm{2}} .\mathrm{2}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${fn}\left({a}\right)={a}^{{n}+\mathrm{2}} \left({I}_{{n}} −{I}_{{n}+\mathrm{2}} \right) \\ $$$${gn}\left({a}\right)=\frac{{d}}{{da}}\int_{\mathrm{0}} ^{{a}} {x}^{{n}} \sqrt{\left({a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}{dx}=\frac{{d}}{{da}}\int_{\mathrm{0}} ^{\mathrm{1}} {a}^{{n}} {t}^{{n}} {a}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}}{{da}}\left({a}^{{n}+\mathrm{2}} {t}^{{n}} \sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right.}\:{dt}\right)= \\ $$$$\left({n}+\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {a}^{{n}+\mathrm{1}} {t}^{{n}} \sqrt{\left(\mathrm{1}−{t}^{\mathrm{2}} \right.}\:{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} {x}^{\mathrm{3}} \sqrt{\left(\mathrm{4}−{x}^{\mathrm{2}} \right)}{dx} \\ $$$${n}=\mathrm{3}\:{a}=\mathrm{2}\: \\ $$$$=\mathrm{2}^{\mathrm{5}} \left({I}\mathrm{1}−{I}\mathrm{3}\right)=\mathrm{2}^{\mathrm{5}} \left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{32}}{\mathrm{3}} \\ $$
Commented by maxmathsup by imad last updated on 31/May/19
$${thanks}\:{sir}. \\ $$