po1ohkaliDrea po1ohkaliDrea
  • 21-11-2016
  • Mathematics
contestada

integrate 1 / sqrt(25 - 16x^2) dx

Respuesta :

Nirina7
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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