Am currently in high school, so I am aware that my mathematical exposure is quite limited.

My current intuition on integration follows the following reasoning:

  1. All functions (to the extent of high school mathematics) can be expressed as the derivative of another function
  2. A derivative and an integral cancel (e.g. $\int\frac{d}{dx}(x+c)dx=x+c$)

u-substitution appears to be used in situations where there is a chain rule involved with the anti-derivative. However, I do not yet see why u-substitution is important.

Consider the following:

\begin{align}\int \sin x (\cos x)^{-\frac{1}{2}}dx\end{align}

While you can substitute $u=\cos x$ and solve, my method of using the anti-derivative directly seems to render u-substitution redundant. In particular:

\begin{align*} \sin x (\cos x)^{-\frac{1}{2}} &= \frac{d}{dx}-(\cos x)^{\frac{1}{2}}+c\\ \int \sin x (\cos x)^{-\frac{1}{2}}dx &= \int \frac{d}{dx}(-(\cos x)^{\frac{1}{2}}+c)dx\\ &=-(\cos x)^{\frac{1}{2}}+c \end{align*}

As u-substitution is used only for composite functions, this also means that the direct substitution of the anti-derivative should suffice as an integrand which can be solved by u-substitution can only be so complex. I will use $\sin(x^2)$ to illustrate this. I will pretend that I am trying to find the anti-derivative though inspection; the only way to preserve the argument $x^2$ is in the following manner:

\begin{align} \sin(x^2)\approx \frac{d}{dx}(-\cos(x^2)) \end{align} But \begin{align} \frac{d}{dx}(-\cos(x^2))=2x\sin(x^2) \end{align} When one tries to add $\frac{1}{2x}$ to $\frac{d}{dx}(-\cos(x^2))$ in order to cancel out the term, the chain rule no longer applies because the derivative becomes a quotient. Thus we conclude that u-substitution cannot be used to integrate $\sin(x^2)$ since the chain rule is not involved with the antiderivative.

As direct substitution of the anti-derivative appears sufficient to resolve integration problems involving u-substitution, in which instances does the method I've outlined fail and u-substitution becomes necessary?

  • $\begingroup$ u substitution is to help you find the integral. Forgive me if I misunderstand, but your method seems to involve guessing the integral, and then using the fundamental theorem of calculus to cancel the integral sign and derivative. Of course, if you can guess any integral, there is no need for any integration methods. $\endgroup$ – Samasambo Sep 1 '16 at 14:35
  • $\begingroup$ This question makes no sense to me whatsoever. $\endgroup$ – The Great Duck Dec 22 '16 at 5:14

What you call u-substitution is never absolutely necessary. It is, like many other tricks used to find integrals, something to help you out. If you see the integral right away, good for you.

You can always show that one function is the integral of another by just derivating it, saying "look, I tried derivating some functions, just for fun, and I found one whose derivative is $x \rightarrow \sin(x)\cos(x)^{-\frac{1}{2}}$". This is a perfectly valid argument.


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