I'm sure this is easy for the most of...

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You are currently reading a thread in /wsr/ - Worksafe Requests

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I'm sure this is easy for the most of you, but i'd just like to know how i get from the first step to the other. Thanks in advance!

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>>40896

Ok OP, this stuff is really simple!

Remember, whenever you take the limit of a constant it simply is the constant. also remember that the limits of two things multiplied is equal to the limit of one * the limit of the other, provided they exist (ex: Lim (X*Y) = LIm X * Lim Y)

Because of this, you can just move allt eh constants to the left, which include 1, -1, and 1/ln3. thusly, you move -1/ ln(3) to the left and then simplify the right (x^2/x has one x cancel, since we arent evaluating at 0)

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it is (-1/ln3) * (x^2/(x))

since -1/ln3 is constant you can take it outside of the limit

now you have x^2/x inside the limit

since inside the limit x is never 0 you can do the simplify x^2/x to x

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>>40902

>>40903

Thank you so much! while we are at this mind to help me with this anti-derivative?

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>>40896

moreover if you were a little shy about canceling an x from the limit, then apply l'Hospital and you'd get (-1/ln(3))* Lim 2x inspected at 0 from the right.

even though it would seem you could then move the 2 over and get a different value (-2/ln3 as opposed to -1/ln3) you'd still notice that lim x is resultant AND the limits for -2/ln3 * lim x at 0+ and -1/ln3 * lim x at 0+ are equal

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>>40904

ah, looks like you may have to do a U sub and then a trig sub.

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>>40904

sqrt(4-9x^4) = 2 * sqrt(1 - ((3/2)x^2)^2)

so either do (3/2)x^2 = u then u = sin t

or directly (3/2)x^2 = sin t

3x dx = cos t dt

x dx = (1/3) cos t dt

x dx / 2 * sqrt(1 - ((3/2)x^2)^2)

becomes

(1/3) cos t dt / (2*sqrt(1-(sin t)'2))

since 1 - (sin t)^2 = (cos t)^2

it becomes

(1/3) cos t dt / (2*cos t)

=(1/6) * dt

anti derivative is (1/6) * t = (1/6) * arcsin((3/2)x^2)

that is if I didn't fuck up too much

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>>40997

I did it earlier and i came to the same result.

I have another one i'm stuck with:

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>>41013

idea is the same = 2 * sqrt(1-(x/2)^2)

x/2 = sint -> dx = 2*cost dt

2 * sqrt(1-(x/2)^2) dx = 2 * cost * 2*cost dt

= 4 * (cos t)^2 dt (using tri identity)

= 4 * (1 + cos 2t)/2 dt = 2(1 + cos 2t)

(after another substitution u = 2t)

so it is 2t + sin2t = 2t + 2*sint*cost

: 2*arcsin(x/2) + 2*(x/2)*sqrt(1-(x/2)^2)

since sint = x/2 you can compute cost using the identity sin^2+cos^2=1

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