Question Number 212339 by Spillover last updated on 10/Oct/24
Answered by Ghisom last updated on 10/Oct/24
![∫_0 ^3 ((x^(3/4) (3−x)^(1/4) )/((5−x)^3 ))dx= [t=(((2x)/(5(3−x))))^(1/4) ; C=((9(2)^(1/4) )/(5(5)^(1/4) ))] =C∫_0 ^∞ (t^6 /((t^4 +1)^3 ))dt= [Ostrogradski′s Method] =(C/(32))[((t^3 (3t^4 −1))/((t^4 +1)^2 ))]_0 ^∞ +((3C)/(32)) ∫_0 ^∞ (t^2 /(t^4 +1))dt= [use common method] =((27π)/(320((10))^(1/4) ))](Q212343.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{{x}^{\mathrm{3}/\mathrm{4}} \left(\mathrm{3}−{x}\right)^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{5}−{x}\right)^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\left(\frac{\mathrm{2}{x}}{\mathrm{5}\left(\mathrm{3}−{x}\right)}\right)^{\mathrm{1}/\mathrm{4}} ;\:{C}=\frac{\mathrm{9}\sqrt[{\mathrm{4}}]{\mathrm{2}}}{\mathrm{5}\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right] \\ $$$$={C}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{\mathrm{6}} }{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\right] \\ $$$$=\frac{{C}}{\mathrm{32}}\left[\frac{{t}^{\mathrm{3}} \left(\mathrm{3}{t}^{\mathrm{4}} −\mathrm{1}\right)}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{3}{C}}{\mathrm{32}}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{use}\:\mathrm{common}\:\mathrm{method}\right] \\ $$$$=\frac{\mathrm{27}\pi}{\mathrm{320}\sqrt[{\mathrm{4}}]{\mathrm{10}}} \\ $$
Answered by Ghisom last updated on 10/Oct/24
$$...\mathrm{or}\:\mathrm{use} \\ $$$${t}=\mathrm{arcsin}\:\frac{\sqrt{\mathrm{6}{x}}}{\mathrm{3}\sqrt{\mathrm{5}−{x}}} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{9}}{\mathrm{5}\sqrt[{\mathrm{4}}]{\mathrm{40}}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{sin}^{\mathrm{5}/\mathrm{2}} \:{t}\:\mathrm{cos}^{\mathrm{3}/\mathrm{2}} \:{t}\:{dt}= \\ $$$$=\frac{\mathrm{9}}{\:\mathrm{10}\sqrt[{\mathrm{4}}]{\mathrm{40}}}\mathrm{B}\:\left(\frac{\mathrm{5}}{\mathrm{4}},\:\frac{\mathrm{7}}{\mathrm{4}}\right)\:=\frac{\mathrm{9}}{\mathrm{10}\sqrt[{\mathrm{4}}]{\mathrm{40}}}×\frac{\mathrm{3}\pi}{\mathrm{16}\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{27}\pi}{\mathrm{320}\sqrt[{\mathrm{4}}]{\mathrm{10}}} \\ $$$$ \\ $$
Answered by MathematicalUser2357 last updated on 12/Oct/24
$$\mathrm{0}.\mathrm{149060872} \\ $$
Commented by Ghisom last updated on 13/Oct/24
$$\mathrm{very}\:\mathrm{accurate}\:\mathrm{you}\:\mathrm{are} \\ $$