Vtyjkyuk

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.

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

$$int sin xcos x,dx$$ 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

$$int fracsqrt xx+1dx$$

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

$$intfraczz^2+12zdz=2intleft(1-frac1z^2+1right)dz.$$

A concrete example of approach 1 may be something like $intfrac11+sqrt x,mathrmdx$ 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 $phi’(z)=2z,mathrmdz$, 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 $int 2xsin x^2,mathrmdx$ 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.

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,

$$(Fcirc phi)'(x)F'(phi(x))phi'(x)=f(phi(x))phi'(x).$$

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 $gcirc phi^-1(x)$ is an antiderivative of $f(x),$ which is what we want. Let's see why: By the chain rule,

$$tag 1(gcirc phi^-1)'(x)= g'(phi^-1(x))cdot(phi^-1)'(x)$$ $$ =f(phi(phi^-1(x)))cdotphi'(phi^-1(x))cdot(phi^-1)'(x).$$

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.