∫9−x2x2dx We will replace the variables:
x=3sint−>dx=3costdt.
∫9−9sin2t27sin2tcostdt=∫31−sin2t27sin2tcostdt=∫cost9sin2tcostdt=9∫sin2tdt
Use transformation:
sin2t=21−cos2t,
then we get
9∫sin2tdt=9∫21−cos2tdt=29∫(1−cos2t)dt=29(t−21sin2t)+C=29(t−sintcost)+C
where C - contact.
t=arcsin3x
sint=3x
cost=1−sin2t=1−9x2
29(t−sintcost)+C=29(arcsin3x−3x1−9x2)+C=29arcsin3x−x9−x2+C
Answer:
∫9−x2x2dx=29arcsin3x−x9−x2+C
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