Difficulties with Substitution Rule (integration)

[quicksilver123]

I can obviously do the chain rule and see how the final expression of the derivative is related to the original function but I can't seem to figure out the substitution Rule as an intuitive way of solving the indefinite integral of functions... bear with me if I'm too verbose, I've attached an image with ,y reasoning...

 

[quicksilver123]

The Lamar university web page on integration offers the same material and proof.

 

[fresh_42]

If $g(x)=v$ then $\int [f(g(x))g'(x)]dx= \int [f(v)v'(x)]dx=\int [f(v)]dv = \int f(u)du$.

Your mistake is $\int [f(v)v'(x)]dx \neq [\int f(v)dv] \cdot [\int v'dv]$.

The essential reason for this is, because a differentiation obeys the Leibniz rule, i.e. leads to a derivative: $d(f\cdot g)= (df)\cdot g + f\cdot (dg),$ and does not allow a factor-wise multiplication: $d(f\cdot g)\neq (df)\cdot (dg)\,.$

 

[quicksilver123]

Could you explicitly explain the correct method in terms of your liebnitz Rule?

 

[fresh_42]

Could you explicitly explain the correct method in terms of your liebnitz Rule?

Sorry, I confused the chain rule and the Leibniz rule. The chain rule corresponds to the substitution rule and the Leibniz rule corresponds to integration by parts. The shortest way to see the equation in (4) is to use the notation with the $d$'s, also called Leibniz notation. Here we get by the substitution $g(x)=u$
$$
\int [f(g(x))g'(x)dx] = \int [f(g(x)) \frac{dg(x)}{dx} dx] = \int [f(g(x))dg(x)] = \int [f(u)du]
$$
where I have only calculated with the terms under the integration. Nothing goes outside as I understood your remark.

You haven't written (in your hand writing) what your integration variable is, so I assume you meant $\int [f(g(x))g'(x)]dx = [\int f(g(x))dx] \cdot [\int u'dx] =[\int f(g(x))dx] \cdot u$. But the integral of a product (and that is why I brought up the Leibniz rule) is given by
$$
[f(g(x))\cdot g(x)]' = f(g(x)) \cdot g'(x) + [f(g(x)]' \cdot g(x) \Longrightarrow \int [f(g(x)) \cdot g'(x)]dx = f(g(x))\cdot g(x) - \int [f(g(x))]' \cdot g(x)dx
$$