Question #9abc1

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Answer 1 · 2017-04-22T20:53:05

$8arcsin(x/4) + 4sin 2arcsin(x/4) + C$

$x=4sint => dx=4costdt$

$t = arcsin(x/4)$

$sqrt(16-x^2) = sqrt(16-(4sint)^2) = sqrt(16-16sin^2t) =$

$= sqrt(16(1-sin^2t)) = sqrt(16cos^2t) = 4cost$

$int sqrt(16-x^2)dx = int 4cost*4costdt = 16 int cos^2tdt =$

$16 int (1+cos2t)/2dt = 8[t+1/2sin2t]+C =$

$= 8arcsin(x/4) + 4sin 2arcsin(x/4) + C$