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Question Number 33619 by NECx last updated on 20/Apr/18
∫x^(5/2) (1−x)^(3/2) dx
$$\int{x}^{\mathrm{5}/\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} {dx} \\ $$
Commented by abdo imad last updated on 20/Apr/18
let put I = ∫ x^(5/2) (1−x)^(3/2) dx .ch.x=sin^2 t give I = ∫ sin^5 x cos^3 x (2sint cost)dt = 2 ∫ cos^4 x sin^6 x dx =2 ∫ (cosx sinx)^4 sin^2 xdx =(1/8) ∫ (sin(2x))^4 ((1−cos(2x))/2)dx =(1/(16)) ∫ ( ((1−cos(4x))/2))^2 (1−cos(2x))dx =(1/(64))∫ (1−2cos(4x) +cos^2 (4x))(1−cos(2x))dx =(1/(64))∫(1−2cos(4x) + ((1+cos(8x))/2))(1−cos(2x))dx =(1/(128))∫ (2 −4cos(4x) +1 +cos(8x))(1−cos(2x))dx =.....until we find the value....
$${let}\:{put}\:{I}\:=\:\int\:\:{x}^{\frac{\mathrm{5}}{\mathrm{2}}} \:\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dx}\:\:.{ch}.{x}={sin}^{\mathrm{2}} {t}\:{give} \\ $$$${I}\:=\:\int\:{sin}^{\mathrm{5}} {x}\:\:{cos}^{\mathrm{3}} {x}\:\left(\mathrm{2}{sint}\:{cost}\right){dt} \\ $$$$=\:\mathrm{2}\:\int\:\:{cos}^{\mathrm{4}} {x}\:{sin}^{\mathrm{6}} {x}\:{dx} \\ $$$$=\mathrm{2}\:\int\:\left({cosx}\:{sinx}\right)^{\mathrm{4}} \:{sin}^{\mathrm{2}} {xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\:\int\:\left({sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{4}} \:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\:\int\:\left(\:\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\:\left(\mathrm{1}−\mathrm{2}{cos}\left(\mathrm{4}{x}\right)\:+{cos}^{\mathrm{2}} \left(\mathrm{4}{x}\right)\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int\left(\mathrm{1}−\mathrm{2}{cos}\left(\mathrm{4}{x}\right)\:+\:\frac{\mathrm{1}+{cos}\left(\mathrm{8}{x}\right)}{\mathrm{2}}\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{128}}\int\:\left(\mathrm{2}\:−\mathrm{4}{cos}\left(\mathrm{4}{x}\right)\:+\mathrm{1}\:+{cos}\left(\mathrm{8}{x}\right)\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=…..{until}\:{we}\:{find}\:{the}\:{value}…. \\ $$
Commented by abdo imad last updated on 20/Apr/18
this method is useful if −1≤x≤1 .
$${this}\:{method}\:{is}\:{useful}\:{if}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:. \\ $$
Commented by NECx last updated on 21/Apr/18
cant it be completed?
$${cant}\:{it}\:{be}\:{completed}? \\ $$$$ \\ $$
Answered by MJS last updated on 21/Apr/18
u=(√x) dx=2(√x)du 2∫u^6 (1−u^2 )^(3/2) du u=sin v du=cos v dv 2∫cos v sin^6 v (1−sin^2 v)^(3/2) dv= =2∫cos^4 v sin^6 v dv=2∫(1−sin^2 v)^2 sin^6 v dv= =2(∫sin^(10) v dv−2∫sin^8 v dv+∫sin^6 v dv) now we have to use ∫sin^n α dα=((n−1)/n)∫sin^(n−2) α dα−((cos α sin^(n−1) α)/n) until we′re done and get ((3v)/(128))+cos v (−(1/5)sin^9 v +((11)/(40))sin^7 v −(1/(80))sin^5 v −(1/(64))sin^3 v −(3/(128))sin v) now we go back... v=arcsin u sin v =u cos v =(√(1−u^2 )) u=(√x) ∫x^(5/2) (1−x)^(3/2) dx= =(3/(128))arcsin (√x)−((√(x(1−x)))/(640))(128x^4 −176x^3 +8x^2 +10x+15)+C
$${u}=\sqrt{{x}} \\ $$$${dx}=\mathrm{2}\sqrt{{x}}{du} \\ $$$$\mathrm{2}\int{u}^{\mathrm{6}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {du} \\ $$$${u}=\mathrm{sin}\:{v} \\ $$$${du}=\mathrm{cos}\:{v}\:{dv} \\ $$$$\mathrm{2}\int\mathrm{cos}\:{v}\:\mathrm{sin}^{\mathrm{6}} \:{v}\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{v}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dv}= \\ $$$$=\mathrm{2}\int\mathrm{cos}^{\mathrm{4}} \:{v}\:\mathrm{sin}^{\mathrm{6}} \:{v}\:{dv}=\mathrm{2}\int\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{v}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{6}} \:{v}\:{dv}= \\ $$$$=\mathrm{2}\left(\int\mathrm{sin}^{\mathrm{10}} \:{v}\:{dv}−\mathrm{2}\int\mathrm{sin}^{\mathrm{8}} \:{v}\:{dv}+\int\mathrm{sin}^{\mathrm{6}} \:{v}\:{dv}\right) \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{use} \\ $$$$\int\mathrm{sin}^{{n}} \:\alpha\:{d}\alpha=\frac{{n}−\mathrm{1}}{{n}}\int\mathrm{sin}^{{n}−\mathrm{2}} \:\alpha\:{d}\alpha−\frac{\mathrm{cos}\:\alpha\:\mathrm{sin}^{{n}−\mathrm{1}} \:\alpha}{{n}} \\ $$$$\mathrm{until}\:\mathrm{we}'\mathrm{re}\:\mathrm{done}\:\mathrm{and}\:\mathrm{get} \\ $$$$\frac{\mathrm{3}{v}}{\mathrm{128}}+\mathrm{cos}\:{v}\:\left(−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{sin}^{\mathrm{9}} \:{v}\:+\frac{\mathrm{11}}{\mathrm{40}}\mathrm{sin}^{\mathrm{7}} \:{v}\:−\frac{\mathrm{1}}{\mathrm{80}}\mathrm{sin}^{\mathrm{5}} \:{v}\:−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{sin}^{\mathrm{3}} \:{v}\:−\frac{\mathrm{3}}{\mathrm{128}}\mathrm{sin}\:{v}\right) \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{go}\:\mathrm{back}… \\ $$$${v}=\mathrm{arcsin}\:{u} \\ $$$$\mathrm{sin}\:{v}\:={u} \\ $$$$\mathrm{cos}\:{v}\:=\sqrt{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$${u}=\sqrt{{x}} \\ $$$$\int{x}^{\mathrm{5}/\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} {dx}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{128}}\mathrm{arcsin}\:\sqrt{{x}}−\frac{\sqrt{{x}\left(\mathrm{1}−{x}\right)}}{\mathrm{640}}\left(\mathrm{128}{x}^{\mathrm{4}} −\mathrm{176}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{15}\right)+{C} \\ $$

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