#How to compute antiderivative in pic?

51 messages · Page 1 of 1 (latest)

glacial axle
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I was thinking of doing the u' times u formula

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but then arctg would have -1 as exponent which makes that formula not applicable

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So what do?

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<@&727457725017096242>

timid cloak
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I recommend using u-sub

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try u=2tan(x)

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things simplify very well

trim inlet
glacial axle
timid cloak
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same thing with sin and sen

trim inlet
glacial axle
timid cloak
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@glacial axle did u try that u-sub?

glacial axle
trim inlet
timid cloak
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oh yh

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I just saw what u have to do

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@glacial axle try computing the derivative of arctg(x/2)

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see what u get

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you'll see a familiar expression

trim inlet
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Holy shit, I'm an idiot. I already did that, but I did it wrong.

glacial axle
# timid cloak you'll see a familiar expression

But if I separate it into two fractions (one multiplied by the other) and make the first fraction be the derivative of the second I still have the problem that the exponent of the second fraction would be -1 which makes the rule invalid.

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You're trying to get to this form, right?

timid cloak
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wdym?

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nop

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there's another form

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lemme write it down

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$\int { \frac{f^{'}(x)}{f(x)} dx} = ln(f(x)) + C$

livid frigateBOT
timid cloak
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it's really the absolute value, but u get the idea

glacial axle
timid cloak
glacial axle
timid cloak
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yes

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now integrate both sides

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do the anti derivative of both sides

glacial axle
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f(x)/int(f(x))

timid cloak
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?

glacial axle
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Not sure what you mean both sides

timid cloak
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u just get the integral of f'(x)/f(x) = ln(f(x) + c

glacial axle
timid cloak
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but still

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u can do this without using it

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but yh, this is the easiest way

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and probably the intended way hahah

hardy bronze
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Let
U = arctan(x/2)*1/2
Du = 1/4 + x^2

$2\int \frac{du}{u} = 2\ln{|u|} + C= 2\ln{|\frac{1}{2}\arctan(\frac{x}{2})|} + C$