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Question Number 99205 by bemath last updated on 19/Jun/20

Commented by bramlex last updated on 19/Jun/20

let (√(4−(√x))) = t , 4−(√x) = t^2 (4−t^2 )^2 = x ; dx = −4t(4−t^2 ) dt I = ∫ t (−4t(4−t^2 )) dt I= −4∫ t^2 (4−t^2 ) dt = −4∫ (4t^2 −t^4 )dt I= −4 {(4/3)t^3 −(1/5)t^5 } + c I = −4t^3 {(1/3)−(1/5)t^2 } + c I= −4(4−(√x))^(3/2) {(1/3)−(1/5)(4−(√x))}+ c

let4−x=t,4−x=t2(4−t2)2=x;dx=−4t(4−t2)dtI=∫t(−4t(4−t2))dtI=−4∫t2(4−t2)dt=−4∫(4t2−t4)dtI=−4{43t3−15t5}+cI=−4t3{13−15t2}+cI=−4(4−x)32{13−15(4−x)}+c

Answered by mathmax by abdo last updated on 19/Jun/20

I =∫(√(4−(√x)))dx we do tbe changement (√(4−(√x)))=t ⇒4−(√x)=t^2 ⇒ (√x)=4−t^2 ⇒x =(4−t^2 )^2 =t^4 −8t^2 +16 ⇒ I =∫ t (4t^3 −16t)dt =∫ (4t^4 −16t^2 )dt =(4/5)t^5 −((16)/3)t^3 +C =(4/5)((√(4−(√x))))^5 −((16)/3)((√(4−(√x))))^3 +C

I=∫4−xdxwedotbechangement4−x=t⇒4−x=t2⇒x=4−t2⇒x=(4−t2)2=t4−8t2+16⇒I=∫t(4t3−16t)dt=∫(4t4−16t2)dt=45t5−163t3+C=45(4−x)5−163(4−x)3+C

Commented by bemath last updated on 19/Jun/20

thank you both

thankyouboth

Commented by mathmax by abdo last updated on 19/Jun/20

you are welcome

youarewelcome

Answered by MJS last updated on 19/Jun/20

∫(√(4−(√x)))dx= [t=(√(4−(√x))) → dx=−4(√x)(√(4−(√x)))dt] x=(t^2 −4)^2 ∧0≤t≤2 ⇔ 0≤x≤16 =4∫t^4 −4t^2 dt=(4/5)t^5 −((16)/3)t^3 = =(4/(15))(3x−4(√x)−32)(√(4−(√x))))+C

∫4−xdx=[t=4−x→dx=−4x4−xdt]x=(t2−4)2∧0⩽t⩽2⇔0⩽x⩽16=4∫t4−4t2dt=45t5−163t3==415(3x−4x−32)4−x)+C

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