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dx-4x-x-2-3-

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Question Number 119784 by bemath last updated on 27/Oct/20
∫ (dx/( (√((4x−x^2 )^3 ))))
∫dx(4x−x2)3
Answered by bobhans last updated on 27/Oct/20
∫ (dx/( (4−(2−x)^2 )^(3/2) )) ; [ let 2−x = 2sin t ] dx = −2cos t dt ] ∫ ((−2cos t dt)/((4−4sin^2 t)^(3/2) )) = ∫ ((−cos t dt)/(4cos^3 t)) =−(1/4)∫ (1/(cos^2 t)) dt = −(1/4)tan t + c =−((2−x)/(4(√(4x−x^2 )))) + c = ((x−2)/(4(√(4x−x^2 )))) + c
∫dx(4−(2−x)2)32;[let2−x=2sint]dx=−2costdt]∫−2costdt(4−4sin2t)32=∫−costdt4cos3t=−14∫1cos2tdt=−14tant+c=−2−x44x−x2+c=x−244x−x2+c
Answered by MJS_new last updated on 28/Oct/20
∫(dx/(x^(3/2) (4−x)^(3/2) ))= [t=((√x)/( (√(4−x)))) → dx=((√(x(4−x)^3 ))/2)dt] =(1/8)∫1+(1/t^2 )dt=(t/8)−(1/(8t))=((x−2)/(4(√(x(4−x)))))+C
∫dxx3/2(4−x)3/2=[t=x4−x→dx=x(4−x)32dt]=18∫1+1t2dt=t8−18t=x−24x(4−x)+C

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