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Question Number 86399 by Power last updated on 28/Mar/20

Commented by Prithwish Sen 1 last updated on 28/Mar/20

put x=2asin^2 t and simplify.

$$\mathrm{put}\:\mathrm{x}=\mathrm{2asin}^{\mathrm{2}} \mathrm{t} \\ $$$$\mathrm{and}\:\mathrm{simplify}. \\ $$

Commented by mathmax by abdo last updated on 28/Mar/20

I =∫ ((x(√x))/(√(2a−x))) dx we do the changement (√(2a−x))=t ⇒2a−x=t^2 ⇒ x=2a−t^2 ⇒ I =∫ (((2a−t^2 )(√(2a−t^2 )))/t)(−2t)dt =−2 ∫ (2a−t^2 )(√(2a−t^2 ))dt =_(t=(√(2a))sinu) −2∫ (2a−2asin^2 u)(√(2a))cosu (√(2a))cosu du =−4a(2a) ∫ (1−sin^2 u)cos^2 u du =−8a^2 ∫ cos^4 u du =−8a^2 ∫ (((1+cos(2u))/2))^2 du =−4a^2 ∫ (1+2cos(2u) +cos^2 (2u))du =−4a^2 { u +sin(2u)}−4a^2 ∫ ((1+cos(4u))/2)du =−4a^2 u −4a^2 sin(2u) −2a^2 (u +(1/4)sin(4u)) +C =−6a^2 u −4a^2 sin(2u)−(a^2 /2)sin(4u)+C =−6a^2 u −8a^2 sinu cosu −a^2 sin(2u)cos(2u) +C =−6a^2 u −8a^2 sinucosu −2a^2 sinu cosu(1−2sin^2 u) +C =−2a^2 { 3 arcsin((t/(√(2a))))+4((t/(√(2a))))(√(1−((t/(√(2a))))^2 )) +((t/(√(2a))))(√(1−((t/(√(2a))))^2 ))(1−2((t/(√(2a))))^2 ) }+C I=−2a^2 {3arcsin(((√(2a−x))/(√(2a))))+((4((√(2a−x))))/(√(2a)))(√(1−(((√(2a−x))/(√(2a))))^2 )) +((√(2a−x))/(√(2a)))(√(1−(((√(2a−x))/(√(2a))))^2 ))(1−2(((√(2a−x))/(√(2a))))^2 } +C

$${I}\:=\int\:\:\frac{{x}\sqrt{{x}}}{\sqrt{\mathrm{2}{a}−{x}}}\:{dx}\:{we}\:{do}\:{the}\:{changement}\:\sqrt{\mathrm{2}{a}−{x}}={t}\:\Rightarrow\mathrm{2}{a}−{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\mathrm{2}{a}−{t}^{\mathrm{2}} \:\Rightarrow\:{I}\:=\int\:\:\frac{\left(\mathrm{2}{a}−{t}^{\mathrm{2}} \right)\sqrt{\mathrm{2}{a}−{t}^{\mathrm{2}} }}{{t}}\left(−\mathrm{2}{t}\right){dt} \\ $$$$=−\mathrm{2}\:\int\:\:\left(\mathrm{2}{a}−{t}^{\mathrm{2}} \right)\sqrt{\mathrm{2}{a}−{t}^{\mathrm{2}} }{dt}\:\:=_{{t}=\sqrt{\mathrm{2}{a}}{sinu}} \:\:−\mathrm{2}\int\:\left(\mathrm{2}{a}−\mathrm{2}{asin}^{\mathrm{2}} {u}\right)\sqrt{\mathrm{2}{a}}{cosu}\:\sqrt{\mathrm{2}{a}}{cosu}\:{du} \\ $$$$=−\mathrm{4}{a}\left(\mathrm{2}{a}\right)\:\int\:\:\left(\mathrm{1}−{sin}^{\mathrm{2}} {u}\right){cos}^{\mathrm{2}} {u}\:{du} \\ $$$$=−\mathrm{8}{a}^{\mathrm{2}} \int\:{cos}^{\mathrm{4}} {u}\:{du}\:=−\mathrm{8}{a}^{\mathrm{2}} \:\int\:\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:{du} \\ $$$$=−\mathrm{4}{a}^{\mathrm{2}} \int\:\:\left(\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}{u}\right)\:+{cos}^{\mathrm{2}} \left(\mathrm{2}{u}\right)\right){du} \\ $$$$=−\mathrm{4}{a}^{\mathrm{2}} \left\{\:\:{u}\:\:+{sin}\left(\mathrm{2}{u}\right)\right\}−\mathrm{4}{a}^{\mathrm{2}} \int\:\:\frac{\mathrm{1}+{cos}\left(\mathrm{4}{u}\right)}{\mathrm{2}}{du} \\ $$$$=−\mathrm{4}{a}^{\mathrm{2}} {u}\:−\mathrm{4}{a}^{\mathrm{2}} \:{sin}\left(\mathrm{2}{u}\right)\:−\mathrm{2}{a}^{\mathrm{2}} \left({u}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{4}{u}\right)\right)\:+{C} \\ $$$$=−\mathrm{6}{a}^{\mathrm{2}} {u}\:−\mathrm{4}{a}^{\mathrm{2}} {sin}\left(\mathrm{2}{u}\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}\left(\mathrm{4}{u}\right)+{C} \\ $$$$=−\mathrm{6}{a}^{\mathrm{2}} {u}\:−\mathrm{8}{a}^{\mathrm{2}} {sinu}\:{cosu}\:−{a}^{\mathrm{2}} \:{sin}\left(\mathrm{2}{u}\right){cos}\left(\mathrm{2}{u}\right)\:+{C} \\ $$$$=−\mathrm{6}{a}^{\mathrm{2}} {u}\:−\mathrm{8}{a}^{\mathrm{2}} {sinucosu}\:−\mathrm{2}{a}^{\mathrm{2}} {sinu}\:{cosu}\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} {u}\right)\:+{C} \\ $$$$=−\mathrm{2}{a}^{\mathrm{2}} \left\{\:\mathrm{3}\:{arcsin}\left(\frac{{t}}{\sqrt{\mathrm{2}{a}}}\right)+\mathrm{4}\left(\frac{{t}}{\sqrt{\mathrm{2}{a}}}\right)\sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}{a}}}\right)^{\mathrm{2}} }\right. \\ $$$$\left.+\left(\frac{{t}}{\sqrt{\mathrm{2}{a}}}\right)\sqrt{\mathrm{1}−\left(\frac{{t}}{\sqrt{\mathrm{2}{a}}}\right)^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{2}\left(\frac{{t}}{\sqrt{\mathrm{2}{a}}}\right)^{\mathrm{2}} \right)\:\right\}+{C} \\ $$$${I}=−\mathrm{2}{a}^{\mathrm{2}} \left\{\mathrm{3}{arcsin}\left(\frac{\sqrt{\mathrm{2}{a}−{x}}}{\sqrt{\mathrm{2}{a}}}\right)+\frac{\mathrm{4}\left(\sqrt{\mathrm{2}{a}−{x}}\right)}{\sqrt{\mathrm{2}{a}}}\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}{a}−{x}}}{\sqrt{\mathrm{2}{a}}}\right)^{\mathrm{2}} }\right. \\ $$$$+\frac{\sqrt{\mathrm{2}{a}−{x}}}{\sqrt{\mathrm{2}{a}}}\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}{a}−{x}}}{\sqrt{\mathrm{2}{a}}}\right)^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{2}\left(\frac{\sqrt{\mathrm{2}{a}−{x}}}{\sqrt{\mathrm{2}{a}}}\right)^{\mathrm{2}} \right\}\:+{C} \\ $$

Commented by jagoll last updated on 28/Mar/20

standard solving let a−x = a cos 2t (x/(2a−x)) = ((1+cos 2t)/(1−cos 2t)) = cot^2 t ∫ 2a^2 ×a(1−cos 2t)sin 2t cot t dt = 8a^2 ∫ sin^2 t cos^2 t dt = 2a^2 ∫ sin^2 2t dt = 2a^2 ∫ ((1/2)−(1/2)cos 4t) dt = a^2 t −(a^2 /4) sin 4t + c super easy to solve

$$\mathrm{standard}\:\mathrm{solving} \\ $$$$\mathrm{let}\:\mathrm{a}−\mathrm{x}\:=\:\mathrm{a}\:\mathrm{cos}\:\mathrm{2t}\: \\ $$$$\frac{\mathrm{x}}{\mathrm{2a}−\mathrm{x}}\:=\:\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2t}}{\mathrm{1}−\mathrm{cos}\:\mathrm{2t}}\:=\:\mathrm{cot}\:^{\mathrm{2}} \mathrm{t}\: \\ $$$$\int\:\mathrm{2a}^{\mathrm{2}} \:×\mathrm{a}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2t}\right)\mathrm{sin}\:\mathrm{2t}\:\mathrm{cot}\:\:\mathrm{t}\:\mathrm{dt} \\ $$$$=\:\mathrm{8a}^{\mathrm{2}} \:\int\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{dt}\: \\ $$$$=\:\mathrm{2a}^{\mathrm{2}} \:\int\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2t}\:\mathrm{dt}\: \\ $$$$=\:\mathrm{2a}^{\mathrm{2}} \:\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4t}\right)\:\mathrm{dt} \\ $$$$=\:\mathrm{a}^{\mathrm{2}} \:\mathrm{t}\:−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{sin}\:\mathrm{4t}\:+\:\mathrm{c}\: \\ $$$$\mathrm{super}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

Answered by john santu last updated on 28/Mar/20

Commented by Power last updated on 28/Mar/20

thank you sir

$$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{sir}} \\ $$

Commented by peter frank last updated on 29/Mar/20

thank you both

$${thank}\:{you}\:{both} \\ $$

Answered by MJS last updated on 28/Mar/20

∫x(√(x/(2a−x)))dx= [t=(√(x/(2a−x))) → dx=((4at)/((t^2 +1)^2 ))dt] =8a^2 ∫(t^4 /((t^2 +1)^3 ))dt= =−((a^2 t(5t^2 +3))/((t^2 +1)^3 ))+3a^2 arctan t = =−(((x+3a)(√(2ax−x^2 )))/2)+3a^2 arctan (√(x/(2a−x))) +C

$$\int{x}\sqrt{\frac{{x}}{\mathrm{2}{a}−{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}}{\mathrm{2}{a}−{x}}}\:\rightarrow\:{dx}=\frac{\mathrm{4}{at}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}\right] \\ $$$$=\mathrm{8}{a}^{\mathrm{2}} \int\frac{{t}^{\mathrm{4}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$=−\frac{{a}^{\mathrm{2}} {t}\left(\mathrm{5}{t}^{\mathrm{2}} +\mathrm{3}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }+\mathrm{3}{a}^{\mathrm{2}} \mathrm{arctan}\:{t}\:= \\ $$$$=−\frac{\left({x}+\mathrm{3}{a}\right)\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }}{\mathrm{2}}+\mathrm{3}{a}^{\mathrm{2}} \mathrm{arctan}\:\sqrt{\frac{{x}}{\mathrm{2}{a}−{x}}}\:+{C} \\ $$

Commented by jagoll last updated on 28/Mar/20

super easy sir?

$$\mathrm{super}\:\mathrm{easy}\:\mathrm{sir}? \\ $$

Commented by MJS last updated on 28/Mar/20

well well well...

$$\mathrm{well}\:\mathrm{well}\:\mathrm{well}... \\ $$

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