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How to integrate it? INT [1/ sqrt(3-t^2) ] dt?
INT [1/ sqrt(3-t^2) ] dt
=INT [1/{sqrt(3) * sqrt(1- (t^2)/3)} dt
=1/sqrt(3) INT [1/( sqrt(1- (t/3^(1/2))^2 ) ] dt
={1/sqrt(3)} arcsin (t/sqrt(3))
:(
2 Answers
- Anonymous1 decade agoFavorite Answer
there's a rule INT [1/ sqrt(a^2 - x^2) dx = sin inverse (x/a) + C
so a= sqrt 3 and x=t
answer: sin inverse (t/sqrt 3) + C
Source(s): http://www.integral-table.com/ - 1 decade ago
antiderivative formula for arcsin (arc means inverse sin^-1)
( au’ )
(aâ(a^2-u^2) ) = arcsin(u/a)
' means derivative
the a's would reduce leaving the derivative of u which is 1 so the antiderivatie is t.
a^2= 3 so a =sqrt3
u^2=t^2 so u =t
arcsin(t/â(3))
Source(s): AP Calculus Class I am taking