is there a reverse chain rule for integrals? if so what is it?

Question

my original problem was integrate (1-x^2)^.5
when differentiating i would use the chain rule but i see that you dont have to use an inverse chain rule for integration in this instance. i realize that the best way to solve is by using

integralA^u du = A^u/lnA +c

so i wonder if a chain rule for integrals exists and what is the proof of integralA^u du = A^u/lnA +c preferably using the product rule for integrals if possible.

Answer 1

Nope, there is no chain rule for integration. There are many simple functions that have no closed form integral. For differentiation, such a case is rare. All you need are a few bags of tricks to differentiate. Integration is almost an art form. The closest thing to a chain rile for integration is integration by parts which is really ust an application of the chain rule.

By the chain rule d(uv) = u dv + v du

If you rearrange it and add integral signs, you get integration by parts.

The specific integration you cite (ntegralA^u du = A^u/lnA + c) arises from the definition of the natural log. If you realize that A^u = e^(u*lnA) and the definiton of e^x requires that (Int)(e^x dx) = e^x
then the the proof of your integration follows directly.

(Int)(A^u du) = (Int)(e^(u*ln A) du) = (1/ln A)*(Int)(e^(u* ln A) d(u * ln A) =(e^(u * ln A))/ln A = (A^u)/ln A

Answer 2

A fundamental theorem of calculus is that

If integral F(x) dx = f(x), then

the derivative of f(x) with respect to x must = F(x)

So to prove your integral is valid...

integralA^u du = A^u/lnA +c

I'll take the right hand end (RHE), differentiate with respect to x and end up with A^u. Keep in mind that lnA is a constant; and I let A^u be e^(lnA * u).

differentiating RHE:

[1/lnA] {e^(u*lnA)} lnA

Where the right end factor and left end factor make 1, and the factor in the middle brace is A^u.
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NOW to your "original problem was integrate (1-x^2)^.5

Mathemat... calls it double angle formula; I call it half angle formula.

Your integral is (1/2) [(arc sin x) + x sqrt{1-x^2}] + k
(and you can verify this using that fundamental theorem I started with).

Answer 3

In 'reverse' of the chain rule is just the method of substitution. Unfortunately, the integral you are doing cannot be done by a simple substitution. Nor is the formula you gave relevant (it requires A to be a constant). The best way to do this one is to let x=sin(theta), which will lead to an integral of cos^2(theta) d(theta). This is usually done via a double angle formula.

Answer 4

There is no 'chain rule' for integration. When integrating you use substitution most of the time.

Integral u^n du = (u^n+1) / (n+1).

Answer 5

Sadly that cannot be applied chain rule is only 1 way and cannot be reversed i wont explain here cause its a bit kinda big