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Integration quick question
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Htx_x346
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#1
How did they integrate this in one step![Name: 2312ED0E-6AFA-4BB4-91EE-4AB1A9036ACC.jpeg
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Last edited by Htx_x346; 4 weeks ago
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mqb2766
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Htx_x346
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#3
(Original post by mqb2766)
Its really just the (reverse) chain rule.
Its really just the (reverse) chain rule.
i was tryna do integration by parts lol
how do you identify when to use reverse chain rule?
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username5960917
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#4
(Original post by Htx_x346)
OHHH
i was tryna do integration by parts lol
how do you identify when to use reverse chain rule?
OHHH
i was tryna do integration by parts lol
how do you identify when to use reverse chain rule?
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tonyiptony
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#5
(Original post by Htx_x346)
OHHH
i was tryna do integration by parts lol
how do you identify when to use reverse chain rule?
OHHH
i was tryna do integration by parts lol
how do you identify when to use reverse chain rule?
But usually when one part of the integrand is the derivative of the other, reverse chain rule (i.e. substitution) is the way.
For instance, for this question, the numerator is the derivative of whatever is inside the square root in the denominator. so letting u="stuff inside square root" is natural.
Sidenote: Try considering substitution before IBP. Integrals requiring IBP often has a function you want to differentiate.
Last edited by tonyiptony; 4 weeks ago
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Htx_x346
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#6
(Original post by tonyiptony)
It comes with practice and recognizing patterns.
But usually when one part of the integrand is the derivative of the other, reverse chain rule (i.e. substitution) is the way.
For instance, for this question, the number is the derivative of whatever is inside the square root in the denominator. so letting u="stuff inside square root" is natural.
Sidenote: Try considering substitution before IBP. Integrals requiring IBP often has a function you want to differentiate.
It comes with practice and recognizing patterns.
But usually when one part of the integrand is the derivative of the other, reverse chain rule (i.e. substitution) is the way.
For instance, for this question, the number is the derivative of whatever is inside the square root in the denominator. so letting u="stuff inside square root" is natural.
Sidenote: Try considering substitution before IBP. Integrals requiring IBP often has a function you want to differentiate.
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tonyiptony
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#7
(Original post by Htx_x346)
Thank you but what do you mean by “often has a function you want to differentiate” ?
Thank you but what do you mean by “often has a function you want to differentiate” ?
.Wouldn't it be nice to differentiate the stray
, just to make it go away? IBP works here.Another one would be
.We don't know how to integrate
, but differentiating it is super easy. IBP also works perfectly here.There are some clues, usually. Like a logarithmic function, an inverse trig function, a stray x^n, that is way more convenient to differentiate; or something like a product of 2 types of functions (e.g.
), etc. Try to do more integrals and see if you can discover some other clues yourself!
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Htx_x346
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#8
(Original post by tonyiptony)
Say the good ol' classic.
Wouldn't it be nice to differentiate the stray, just to make it go away? IBP works here.
Another one would be.
We don't know how to integrate, but differentiating it is super easy. IBP also works perfectly here.
There are some clues, usually. Like a logarithmic function, an inverse trig function, a stray x^n, that is way more convenient to differentiate; or something like a product of 2 types of functions (e.g.), etc. Try to do more integrals and see if you can discover some other clues yourself!
Say the good ol' classic
.Wouldn't it be nice to differentiate the stray
, just to make it go away? IBP works here.Another one would be
.We don't know how to integrate
, but differentiating it is super easy. IBP also works perfectly here.There are some clues, usually. Like a logarithmic function, an inverse trig function, a stray x^n, that is way more convenient to differentiate; or something like a product of 2 types of functions (e.g.
), etc. Try to do more integrals and see if you can discover some other clues yourself!
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tonyiptony
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#9
(Original post by Htx_x346)
For the first example wouldn’t it just be easier to use reverse chain rule?
For the first example wouldn’t it just be easier to use reverse chain rule?
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mqb2766
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#10
(Original post by wiseowlz72)
f'(x)/f(x)
f'(x)/f(x)
x * (4-x^2)^(-1/2)
Then you can spot that its of the form
k * du/dx * u^(-1/2)
where u = 4-x^2 and k is a constant. So it integrates to something like
u^(1/2)
Last edited by mqb2766; 4 weeks ago
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