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

$$\frac{⊚\mathbb{B}e\mathcal{M}ath⊚}{}$$$$\underset{0}{\stackrel{1}{\int }}{x}^{\frac{9}{2}}{(1-x)}^{\frac{5}{2}}dx?$$

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

$$\mathrm{Set}\frac{1}{\mathrm{x}}-1={\mathrm{t}}^2\Rightarrow -\frac{1}{{\mathrm{x}}^2}\mathrm{dx}=2\mathrm{tdt}\Rightarrow \mathrm{dx}=-{2\mathrm{tx}}^2\mathrm{dt}$$$$\mathrm{t}(\mathrm{x}=1)=0,\mathrm{t}(\mathrm{x}=0)=\mathrm{\infty }$$$$\mathrm{x}=\frac{1}{\sqrt{1+{\mathrm{t}}^2}}\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\frac{9}{2}}{\left({\mathrm{xt}}^2\right)}^{\frac{1}{2}}{2\mathrm{tx}}^2\mathrm{dt}$$$$=2{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^7{\mathrm{t}}^2\mathrm{dt}=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^2\mathrm{dt}}{\sqrt{{(1+{\mathrm{t}}^2)}^7}}$$$$=2{\int }_1^{\mathrm{\infty }}\frac{{\mathrm{t}}^2+1-1}{{(1+{\mathrm{t}}^2)}^3\sqrt{1+{\mathrm{t}}^2}}\mathrm{dt}=2\int \frac{\mathrm{dt}}{{(1+{\mathrm{t}}^2)}^2\sqrt{1+{\mathrm{t}}^2}}$$$$-2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{(1+{\mathrm{t}}^2)}^3\sqrt{1+{\mathrm{t}}^2}}=$$$$\mathrm{Set}\mathrm{t}=\tan \phi \Rightarrow \mathrm{dt}=(1+{\mathrm{t}}^2)\mathrm{d}\phi$$$$\mathrm{I}=2{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{d}\phi }{(1+{\tan }^2\phi )\sqrt{1+{\tan }^2\phi )}}$$$$-2{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{d}\phi }{{(1+{\tan }^2\phi )}^2\sqrt{1+{\tan }^2\phi )}}$$$$=2{\int }_0^{\frac{\pi }{2}}{\cos }^3\phi \mathrm{d}\phi -2{\int }_0^{\frac{\pi }{2}}{\cos }^5\mathrm{d}\phi$$$$=2{\int }_0^{\frac{\pi }{2}}(1-{\sin }^2\phi )\mathrm{dsin}\phi -2{\int }_0^{\frac{\pi }{2}}{(1-{\sin }^2\phi )}^2\mathrm{dsin}\phi$$$$=(2\sin \phi -\frac{{2\sin }^3\phi }{3}){\mid }_0^{\frac{\pi }{2}}-2{\int }_0^{\frac{\pi }{2}}[1-{2\sin }^2\phi +{\sin }^4\phi )\mathrm{dsin}\phi$$$$=2-\frac{2}{3}-2{[\sin \phi -\frac{2}{3}{\sin }^3\phi +\frac{{\sin }^5\phi }{5}]}_0^{\frac{\pi }{2}}$$$$=2-\frac{2}{3}-2(1-\frac{2}{3}+\frac{1}{5})=\frac{4}{15}$$

Answered by bobhans last updated on 13/Aug/20

$$\frac{\nparallel \mathcal{B}\mathrm{ob}\mathcal{H}\mathrm{ans}\nparallel }{\mathrm{\Pi }}$$$$\mathrm{I}=-\frac{2}{7}\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^{9/2}\mathrm{d}{(1-\mathrm{x})}^{7/2}=-\frac{2}{7}{\left[{\mathrm{x}}^{9/2}{(1-\mathrm{x})}^{7/2}\right]}_0^1+$$$$\frac{9}{7}\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^{7/2}{(1-\mathrm{x})}^{7/2}\mathrm{dx}$$$$\mathrm{let}\mathrm{J}=\underset{0}{\stackrel{1}{\int }}{\mathrm{x}}^{7/2}{(1-\mathrm{x})}^{7/2}\mathrm{dx}=\frac{9}{7}\underset{0}{\stackrel{1}{\int }}{(\mathrm{x}-{\mathrm{x}}^2)}^{7/2}\mathrm{dx}$$$$\mathrm{J}=\frac{9}{7}\underset{0}{\stackrel{1}{\int }}{[\frac{1}{4}-{(\frac{1}{2}-\mathrm{x})}^2]}^{7/2}\mathrm{dx}$$$$\mathrm{set}\frac{1}{2}-\mathrm{x}=\mathrm{z}$$$$\mathrm{J}=\frac{9}{7\times 2^8}\underset{-\pi /2}{\stackrel{\pi /2}{\int }}\cos {}^8\mathrm{z}\mathrm{dz}$$$$\mathrm{Reduction}\mathrm{formula}$$$${\mathrm{J}}_{\mathrm{n}}=\frac{\mathrm{n}-1}{\mathrm{n}}.{\mathrm{J}}_{\mathrm{n}-2}\Rightarrow {\mathrm{J}}_8=\frac{7}{8}.\frac{5}{6}.\frac{3}{4}.\frac{1}{2}.{\mathrm{J}}_0$$$${\mathrm{J}}_8=\frac{35}{128}.[\frac{\pi }{2}-(-\frac{\pi }{2})]=\frac{35\pi }{128}$$$$\mathrm{conclusion}$$$$\mathrm{I}=\frac{9}{7\times 2^8}\times \frac{35\pi }{128}=\frac{45\pi }{32768}.$$$$$$

Commented by bemath last updated on 14/Aug/20

$$correct....thankyou$$

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

$$\mathrm{I}={\int }_0^1{\mathrm{x}}^{\frac{9}{2}}{(1-\mathrm{x})}^{\frac{5}{2}}\mathrm{dx}\mathrm{we}\mathrm{know}\mathrm{B}(\mathrm{p},\mathrm{q})={\int }_0^1{\mathrm{x}}^{\mathrm{p}-1}{(1-\mathrm{x})}^{\mathrm{q}-1}\mathrm{dx}$$$$=\frac{\mathrm{\Gamma }\left(\mathrm{p}\right).\mathrm{\Gamma }\left(\mathrm{q}\right)}{\mathrm{\Gamma }(\mathrm{p}+\mathrm{q})}\mathrm{here}\mathrm{p}=\frac{9}{2}+1\mathrm{and}\mathrm{q}=\frac{5}{2}+1\Rightarrow$$$$\mathrm{I}=\frac{\mathrm{\Gamma }(\frac{9}{2}+1)\mathrm{\Gamma }(\frac{5}{2}+1)}{\mathrm{\Gamma }(\frac{9}{2}+\frac{5}{2}+2)}=\frac{9}{2}.\frac{5}{2}.\frac{\mathrm{\Gamma }\left(\frac{9}{2}\right)\mathrm{\Gamma }\left(\frac{5}{2}\right)}{\mathrm{\Gamma }\left(9\right)}=\frac{45}{4.8!}\mathrm{\Gamma }\left(\frac{9}{2}\right)\mathrm{\Gamma }\left(\frac{5}{2}\right)$$$$\mathrm{\Gamma }\left(\frac{9}{2}\right)=\mathrm{\Gamma }(1+\frac{7}{2})=\frac{7}{2}\mathrm{\Gamma }\left(\frac{7}{2}\right)=\frac{7}{2}\mathrm{\Gamma }(1+\frac{5}{2})=\frac{7}{2}.\frac{5}{2}\mathrm{\Gamma }\left(\frac{5}{2}\right)$$$$=\frac{35}{4}\mathrm{\Gamma }(1+\frac{3}{2})=\frac{35}{4}.\frac{3}{2}\mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{3.35}{8}\mathrm{\Gamma }(\frac{1}{2}+1)=\frac{3.35}{16}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$\mathrm{we}\mathrm{follow}\mathrm{the}\mathrm{same}\mathrm{way}\mathrm{to}\mathrm{find}\mathrm{\Gamma }\left(\frac{5}{2}\right)$$

Commented by bemath last updated on 14/Aug/20

$$typoitshouldbe8!indenumerator$$

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