Understanding the integral of $\frac{dy}{dx}$ with respect to $y$ requires a shift in perspective from standard single-variable calculus. Because of that, it represents a distinct mathematical operation that cannot be simplified by simply "canceling" the $dy$ terms. Even so, in typical introductory courses, students become comfortable integrating derivatives with respect to their independent variable—specifically, $\int \frac{dy}{dx} , dx = y + C$. Still, the expression $\int \frac{dy}{dx} , dy$ appears frequently in advanced physics, differential equations, and variational calculus. This article explores the meaning, the rigorous methods for evaluation, and the practical applications of this often-misunderstood integral.
The Notational Trap: Why You Cannot "Cancel" Differentials
The most common error when encountering $\int \frac{dy}{dx} , dy$ is treating Leibniz notation as a simple fraction. A student might be tempted to write:
$ \int \frac{dy}{dx} , dy = \int \frac{dy \cdot dy}{dx} = \int \frac{(dy)^2}{dx} $
Or, even more incorrectly, they might attempt to "cancel" the $dy$ in the numerator of the derivative with the $dy$ of the differential, yielding $\int \frac{1}{dx}$, which is meaningless in standard calculus.
Leibniz notation $\frac{dy}{dx}$ is a single symbol representing the limit of a difference quotient; it is not a ratio of two independent infinitesimals $dy$ and $dx$. While non-standard analysis provides a framework where infinitesimals exist rigorously, in standard real analysis, $dy$ and $dx$ only have meaning inside an integral sign or as differential forms. Which means, algebraic cancellation is invalid. To evaluate this integral correctly, we must use the Chain Rule or Integration by Substitution to transform the variable of integration into a consistent framework.
Method 1: Substitution via the Chain Rule (The Standard Approach)
The most strong way to evaluate $\int \frac{dy}{dx} , dy$ is to change the variable of integration from $y$ to $x$. Since $y$ is a function of $x$ (denoted $y = f(x)$), the differential $dy$ is related to $dx$ by the derivative:
$ dy = \frac{dy}{dx} , dx = f'(x) , dx $
Substitute this expression for $dy$ directly into the integral:
$ \int \frac{dy}{dx} , dy = \int \frac{dy}{dx} \left( \frac{dy}{dx} , dx \right) $
Since $\frac{dy}{dx}$ is a function of $x$, this simplifies to:
$ \int \left( \frac{dy}{dx} \right)^2 , dx $
Result: The integral of $\frac{dy}{dx}$ with respect to $y$ is equivalent to the integral of the square of the derivative with respect to $x$ Easy to understand, harder to ignore..
$ \boxed{\int \frac{dy}{dx} , dy = \int \left( \frac{dy}{dx} \right)^2 , dx} $
This transformation is the cornerstone of solving these problems. It converts an integral with a "mismatched" differential into a standard integral with respect to the independent variable $x$.
Method 2: Inverting the Derivative (When $x$ is a Function of $y$)
If the relationship is given as $x = g(y)$ (where $x$ is the dependent variable), it is often easier to invert the derivative first. Recall the Inverse Function Theorem:
$ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \quad \text{provided} \quad \frac{dx}{dy} \neq 0 $
Substitute this into the original integral:
$ \int \frac{dy}{dx} , dy = \int \frac{1}{\frac{dx}{dy}} , dy $
Now the integrand is expressed purely in terms of $y$ (since $\frac{dx}{dy}$ is the derivative of $x$ with respect to $y$), and the differential $dy$ matches the independent variable. This form is directly integrable with respect to $y$.
$ \boxed{\int \frac{dy}{dx} , dy = \int \frac{1}{x'(y)} , dy} $
This method is particularly powerful when $x(y)$ is simpler than $y(x)$, or when the derivative $\frac{dx}{dy}$ is a basic algebraic function.
Worked Examples: Applying the Theory
Example 1: Polynomial Function ($y = x^3$)
Evaluate $\int \frac{dy}{dx} , dy$ for $y = x^3$.
Step 1: Find the derivative. $ \frac{dy}{dx} = 3x^2 $
Step 2: Apply Method 1 (Substitution). $ \int \frac{dy}{dx} , dy = \int (3x^2)^2 , dx = \int 9x^4 , dx $
Step 3: Integrate with respect to $x$. $ = 9 \frac{x^5}{5} + C = \frac{9}{5}x^5 + C $
Verification via Method 2: Express $x$ in terms of $y$: $x = y^{1/3}$. Find $\frac{dx}{dy} = \frac{1}{3}y^{-2/3}$. The integrand becomes $\frac{1}{\frac{1}{3}y^{-2/3}} = 3y^{2/3}$. Integrate with respect to $y$: $ \int 3y^{2/3} , dy = 3 \frac{y^{5/3}}{5/3} + C = \frac{9}{5} y^{5/3} + C $ Substitute back $y = x^3$: $ \frac{9}{5} (x^3)^{5/3} + C = \frac{9}{5}x^5 + C $ Both methods yield the identical result.
Example 2: Trigonometric Function ($y = \sin x$)
Evaluate $\int \frac{dy}{dx} , dy$ for $y = \sin x$.
Method 1: $ \frac{dy}{dx} = \cos x $ $ \int \cos x \cdot \cos x , dx = \int \cos^2 x , dx $ Use the power-reduction identity $\cos^2 x = \frac{1 + \cos 2x}{2}$: $ \int \frac{1 + \cos 2x}{2} , dx = \frac{1}{2}x + \frac{1}{4}\sin 2x + C $ Using $\sin 2x = 2\sin x \cos x$ and $y = \sin x$: $ = \frac{1}{2}x + \frac{1}{2}\sin x \cos x + C = \frac{1}{2}(x + y\sqrt{1-y^2}) + C $
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