# Two approaches to the method of integration by substitution

I came across two approaches to the method of integration by substitution (in two different books).

Approach I

Let $I=\int f(\phi(x))\phi'(x) dx$

Let $z=\phi(x)$

$\therefore \phi'(x)dx=dz$

$\therefore I=\int f(z)dz$

Approach II

Let $I=\int f(x) dx$

Let $x=\phi(z)$

$dx=\phi'(z) dz$

$\therefore I=\int f(\phi(z))\phi'(z) dz$

My problem: While i can understand Approach I, I cannot understand Approach II. What is the difference between the two approaches. What is the difference in their applicability and usage? I am very confused. Please help.

• Approach 1 looks wrong to me. If $x = \phi(z)$ then $dx$ should equal $\phi'(z)\,dz$. – johnnyb Apr 27 '18 at 16:59
• @johnnyb, that was a typing mistake.I have edited the question. – MrAP Apr 27 '18 at 17:07
• this "identity" $dx=\frac{dz}{\phi'(x)}$ cannot be right, because $\phi'$ could be zero for any point of it domain. Also state change of variables in indefinite integral is not the best way to understand it because indefinite integrals are not, in general, well-defined. – Masacroso Apr 30 '18 at 10:38
• @Masacroso, I have edited that part. Please take a look at it now. – MrAP Apr 30 '18 at 11:05
• It seems that your approach 2 reverses approach 1. So aren't they the same, but than in different direction? Btw, from my experience you use approach 2 in practice. – Jens Wagemaker Apr 30 '18 at 18:02

The two approaches are the same, but one taken forward and the other backward.

The first form is used when the factor $\phi'(x)$ seems obvious.

For instance, in

$$\int \sin x\cos x\,dx$$ you can use $\cos x=\sin'x$ and the integral becomes

$$\int z\,dz.$$

The second form is used when you hope that $f(\phi(z))$ will be simpler than $f(x)$.

For instance, you want to get rid of the square root in

$$\int \frac{\sqrt x}{x+1}dx$$

with the subsitution $x=\phi(z)=z^2$, giving

$$\int\frac{z}{z^2+1}2zdz=2\int\left(1-\frac1{z^2+1}\right)dz.$$

A concrete example of approach 1 may be something like $\int\frac{1}{1+\sqrt x}\,\mathrm{d}x$ and you make the substitution $x=z^2$ in order to get rid of the square root. In this case our $\phi(z)=z^2$ and $\phi’(z)=2z\,\mathrm{d}z$, this makes our integral solvable by some trivial algebra and is already completely in terms of $z$ without any extra algebraic manipulation. Approach 2 on the other hand noticed that there is a derivative of a function on the outside such as $\int 2x\sin x^2\,\mathrm{d}x$ and one makes the substitution $z=x^2$. Both of these are ways to reverse the chain rule as you may recall $(f(g(x)))’=f’(g(x))g’(x)$, although the second approach is pretty much explicitly reversing the chain rule so is the first one in a different manner.

• How do you understand where to use Approach I and where to use Approach II? – MrAP May 5 '18 at 18:22
• It’s on a case to case basis with these types of substitutions. Approach 1 is used commonly for like when you see a derivative on the outside whereas approach 2 is using a substitution to perhaps get a better idea of what the next step should be – Teh Rod May 5 '18 at 18:27

Neither approach as you have it is quite right, because each is missing the final step. In approach 1, suppose you can find an antiderivative $F(z)$ for $f(z).$ Are you done? No, remember you were looking for an antiderivative for $f(\phi(x))\phi'(x).$ The desired answer is $F(\phi(x)).$ Why? Because by the chain rule,

$$(F\circ \phi)'(x)F'(\phi(x))\phi'(x)=f(\phi(x))\phi'(x).$$

Approach 2 also follows from the chain rule, but it's more complicated. Here it's important that $\phi^{-1}$ exist and be differentiable. If we then find an antiderivative $g(z)$ for $f(\phi(z))\phi'(z),$ then $g\circ \phi^{-1}(x)$ is an antiderivative of $f(x),$ which is what we want. Let's see why: By the chain rule,

$$\tag 1(g\circ \phi^{-1})'(x)= g'(\phi^{-1}(x))\cdot(\phi^{-1})'(x)$$ $$=f(\phi(\phi^{-1}(x)))\cdot\phi'(\phi^{-1}(x))\cdot(\phi^{-1})'(x).$$

Recalling $(\phi^{-1})'(x)= 1/[\phi'(\phi^{-1}(x))]$ (again by the chain rule), we see the last two factors on the right of $(1)$ cancel, leaving us with $f(\phi(\phi^{-1}(x))) = f(x)$ as desired.