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find-arcsin-2x-1-4x-2-dx-




Question Number 45970 by maxmathsup by imad last updated on 19/Oct/18
find ∫  ((arcsin(2x))/( (√(1−4x^2 ))))dx
findarcsin(2x)14x2dx
Commented by maxmathsup by imad last updated on 20/Oct/18
changement arcsin(2x)=t  give  2x=sint ⇒2dx=cost dt ⇒  ∫ ((arcsin(2x))/( (√(1−4x^2 ))))dx = (1/2)∫  (t/(cost)) cost dt =(1/2) ∫ t dt =(t^2 /4) +c   =(1/4)(arcsin(2x))^2  +c .
changementarcsin(2x)=tgive2x=sint2dx=costdtarcsin(2x)14x2dx=12tcostcostdt=12tdt=t24+c=14(arcsin(2x))2+c.
Answered by  last updated on 19/Oct/18
∵sin^(−1) 2x=t  (1/( (√(1−4x^2 ))))dx=(1/2)dt  =(1/2)∫tdt  =(1/4)[sin^(−1) 2x]^2 +c
sin12x=t114x2dx=12dt=12tdt=14[sin12x]2+c

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