po1ohkaliDrea
po1ohkaliDrea
21-11-2016
Mathematics
contestada
integrate 1 / sqrt(25 - 16x^2) dx
Respuesta :
Nirina7
Nirina7
22-11-2016
the main formula is integr U^p du = (1/1+p)U^p+1
let be U=
(25 - 16x^2)^-1/2,
1/sqrt(25 - 16x^2) = (25 - 16x^2)^-1/2
du= -1/2)(-32x)(25 - 16x^2)^-1/2dx
integ du/-1/2)(-32x)= integ(25 - 16x^2)^-1/2dx
nteg du/16x)=1/16xu+c=integ(25 - 16x^2)^-1/2dx
finally
integ 1 / sqrt(25 - 16x^2) dx = 1/16xu+c= (1/16x)(
(25 - 16x^2)^-1/2) +c
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