I've made this post both to see if I'm thinking right and to let others read and understand where the "u-substitution" method for integration comes from. I really hate substitutions, because you lost track of what's happening. I've read the related posts in this forum and concluded the following:

The integral u-substitution is a nice method to find some integrals. It comes from the chain rule:

$$\frac{df(g(x))}{dx} = \frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}$$

For me, $\frac{df(x)}{dx}$ is just a notation for the derivative of the function $f$ with respect to $x$, so there's no mean for just $df$ or just $dx$ alone.

When we integrate both sides:

$$\int \frac{df(g(x))}{dx}dx = \int\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}dx$$

Then: $$\underbrace{f(g(x)) + C}_{\text{integral of a derivative}}= \int\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}dx\tag{1}$$

So if we want to integrate some function in the form $\int\frac{df(g(x))}{dg(x)}\frac{dg(x)}{dx}dx$ this is gonna be equal $f(g(x)) + C$. That's why we can integrate $\cos(2x)$ this way:

$$\int \cos(2x)dx = \int \frac{d\sin(2x)}{d2x}\frac{2}{2}dx = \frac{1}{2}\int\frac{d\sin(2x)}{d2x}\cdot2 \ \ dx$$

See how I didn't change the integrand at all, but I multiplied and divided by $2$ to get the form $$\frac{d\sin(2x)}{d[2x]}\frac{d[2x]}{dx} = \frac{d\cos(2x)}{dx}$$ Then, I can match the pattern in $(1)$ to integrate like this: $$\begin{align} &\int\frac{d\color{#F01C2C}{f(}\color{Blue}{g(x)}\color{#F01C2C}{)}}{d\color{Blue}{g(x)}}\color{#01cf84}{\frac{dg(x)}{dx}}dx = \color{#F01C2C}{f(}\color{Blue}{g(x)}\color{#F01C2C}{)} + C \\ \int \cos(2x)dx = \frac{1}{2}&\int\frac{d\color{#F01C2C}{\sin(\color{Blue}{2x})}}{d\color{Blue}{2x}}\cdot\color{#01cf84}{\ \ 2} \ \ \ \ dx = \frac{1}{2}\color{#F01C2C}{\sin(\color{Blue}{2x})} + C\end{align}$$

So... am I right? Do you feel comfortable doing substitutions? Would this technique be acceptable in my math tests? (I really prefer this than the substitution method).


Let's do this integral: $$\int x\ln(\cos(x^2))\sin(x^2)\mathrm dx$$ I will derivate $\cos(x^2)$:

$$\frac{d}{dx}\cos(x^2) = -2x\sin(x^2)$$

Then I'll multiply and divide the integrand by this result:

$$ \begin{align} \int x\ln(\cos(x^2))\sin(x^2) \color{#F01C2C}{\frac{-2x\sin(x^2)}{-2x\sin(x^2)}}dx = \color{#F01C2C}{-\frac{1}{2}}&\int \ln(\cos(x^2))\cdot\color{#F01C2C}{-2x\sin(x^2)}dx \\ &\int\frac{df(g(x))}{dg(x)} \ \ \ \ \ \ \frac{dg(x)}{dx} \ \ \ \ \ \ \ dx \\=&f(g(x)) + C \end{align} $$ So to integrate this, we just have to find the antiderivative of $\ln$ and apply it to the 'point' $\cos(x^2)$. The antiderivative of $\ln$ is $x(\ln(x) - 1)$ by integration by parts. Applying it to $\cos(x^2)$ we have: $\cos(x^2)(\ln(\cos(x^2))-1)$ (this is the antiderivative at $\cos(x^2)$ or $g(x)$. Back in our integral:

$$\color{#F01C2C}{-\frac{1}{2}}\int \ln(\cos(x^2))\cdot\color{#F01C2C}{-2x\sin(x^2)}dx = \color{#F01C2C}{-\frac{1}{2}}\cos(x^2)(\ln(\cos(x^2))-1)$$

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    $\begingroup$ Just something to think about: How would you apply that method to an integral such as $$\int x\ln(\cos(x^2))\sin(x^2)\mathrm dx?$$ $\endgroup$ – homegrown May 24 '14 at 23:23
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    $\begingroup$ Oh, I see. Well, I think this is normal when you polish a method for practical use. Compare integration by parts: the usual formula obscures the connection to the product rule, but is well-adapted to the practical situation where you have one integral and want to change it into a more tractable one. Of course it's good to be familiar with the origin of a technique, but that's usually just a distraction when you're actually computing. $\endgroup$ – user21467 May 25 '14 at 0:22
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    $\begingroup$ @LucasZanella That is the correct answer. But you could have done it like this. Let $u=\cos(x^2),$ then $du=-2x\sin(x^2)\ dx.$ Then we have $-\frac12\int\ln(u)\mathrm du.$ Since you know how to find the antiderivative of $-\frac12\ln u$, then you know it will be $-\frac{u}{2}(\ln u-1).$ Now just plug substitute it back. This just seems shorter than your method. But, you do have the correct answer. $\endgroup$ – homegrown May 25 '14 at 0:25
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    $\begingroup$ I'm doing all this because I'm thinking in how a computer could integrate by some algorithm, and it seems that if I understand the reverse chain rule, I can make a better algorithm. Also, how do you guys determine the right function to substitute for $u$? I only know wich one to do, because I always try to find the derivative of the inside function, in the integrand. $\endgroup$ – Lucas Zanella May 25 '14 at 0:29
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    $\begingroup$ @Neal thanks! I like to think and then ask, nobody's here to do the other's homework. The technique you mentioned is exactly how I think in my method. Thanks again :) $\endgroup$ – Lucas Zanella May 25 '14 at 0:33

By the comments and the question itself, you seem to be looking for techniques to get integrals (antiderivatives, really) symbolically. There are algorithms for this (see e.g. symbolic integration), but they aren't very suitable for hand calculation.


To answer your question: Yes, I feel comfortable with the method of "$u$-substitution". I don't feel comfortable with calling it "$u$-substitution" because it seems silly to name it after a letter of the alphabet. (I don't like the term "$M$-test" very much either.) I just call it "substitution", so I can use whatever letter I like for the new variable.

To answer your other question: If I were teaching calculus, I would consider it quite acceptable for you to use your technique on a test. However, if you were to complain that the test was too long, I might point out that I'd shown you a more efficient procedure for such problems.


I always felt that the substitution method is used only to 'simplify the integral', nothing more. Though substitution method makes you lose track of what you were doing, all you need to do is to organize the way you write.

Though your way to solve the integral is quite impressive, I feel that one would lose track during the start of the race by following your method.

Consider an integral $$ \int_{}^{} [ \sqrt{cot x} + \sqrt{tan x} ]dx $$

where you need to substitute twice, which would be a way more problematic integral if you do not use substitution. Its just like pouring fuel to fire.

I am not saying your method sucks, (damn, I never thought of solving it that way! You are too good!) but just that substitution method provides you more flexibility without writing much.


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