Integrals 3: Our Simple Function

Integration By Parts

Integrals are everywhere in physics, and over and over again the method employed is Integration By Parts. The solution to an integral can be solved with the IBP if the integrand can be broken into two multiplicative factors, using the following method.

When the derivative of a product function, $F'=(UV)'$, F-prime equals u-v-all-primed, is integrated you get the function plus a constant, and the product is worked out.

$$ \begin{equation}\label{FTC} \int F' \, dx = \int (UV' + U'V) \end{equation} $$

We can look at the two terms of a product function's derivative and use the identity created by applying the Product Rule to the product function. The Product Rule and the Fundamental Theorem of Calculus, eq. \eqref{FTC}, is all we need for IBP. The article section on it can be reviewed here: Product Rule .

$$ \begin{equation} \label{IBP} \int (UV)' \, dx = \int U'V \, dx + \int UV' \, dx = UV + \text{constant} \end{equation} $$

The way the Integration By Parts formula, eq. \eqref{IBP}, is used is by decomposing an integral in question into one of the two additive integrals, which then makes the job of integration simpler because the decomposition separated the integration into a formula of two smaller integrals, with one factor always working best as the derivative factor (primed factor).

Integrating Our Simple Function

The antiderivative of a function is generally not obvious, and our example function, that of the normalized one-half power binomial, $f(x)=\sqrt{1+x^2}$, is no exception.

$$ \begin{equation}\label{simple_integral} F = \int_a^y \sqrt{1+x^2} \, dx \end{equation} $$

Since the function has an inner-function, $x^2$, you can't just use the Power Rule on the outer function, $\sqrt{x}$, because the Chain Rule requires us to account for the derivative of the inner function, $1+x^2$. Next, IBP, eq. \eqref{IBP}, can't simply be applied in the current form because $f$ is not a function of a product, it's a single composite function. To solve for $F$, the integral of $f$, eq. \eqref{simple_integral}, we must change its form with a change of variable.

Change of Variable

Since we have a composite function in the integrand, we must use $u$-substitution. The letter u chosen is reasonably thought of as short for Unkown intermediate function. $u$-substitution is the process of transforming the domain, and consequently the form of the integrand in the new domain's variable is different—and amenable to the by parts integration method described above. This change of variable is called a Point Transformation (as opposed to function transformations which finds an equivalent form of the whole function).

Let's consider the tangent function on its principal branch.

Figure 1. Tangent around the origin, on the interval $[-\pi/2 + 0.071, \pi/2 - 0.071]$.

The tangent function is a valid substitution for $x$ because:

their ranges are identical: $[-\infty, \infty]$,
tangent is continuous and differentiable on the principal interval, $[-\pi/2, \pi/2]$,
it is one-to-one (a.k.a, invertible) on this interval.

This invertibility ensures a unique mapping, or bijection, between the variable $x$ and the angle $\theta$. If two functions map separate domains to the same range via continuous bijections, they are compatible substitutes.

Changing the variable has consequences for the integration beyond the composite function. Recall that the width of the infinitesimal slice of the area under the curve in inductive figure is some width in space $x$, but $dx$ transforms as its own function into the new slice transformation, $g(\theta)d\theta$, determined by the derivative of the point transformation, $dx/d\theta$.

Let $x$, along with integration endpoints $a$ and $y$, be substituted out by the tangent function, on its principal branch. The endpoints are transformed with the substitution equivalency, itself.

$$ \begin{equation}\label{simple_integral_endpoints} \theta_0 = \arctan(a)\quad , \quad \theta_1 = \arctan(y) \end{equation} $$

The infinitesimal unit, $dx$, is transformed with an operation that is the same for every type of integration.

$$ \begin{equation}\label{simple_integral_inf} \frac{dx}{d\theta} = \frac{d}{d\theta}\tan(\theta) = 1 + \tan^2(\theta) = \sec^2(\theta) \end{equation} $$

With the substitution of infinitesimals being the following:

$$ \begin{equation}\label{simple_integral_inf_sub} dx = \sec^2(\theta) \, d\theta \end{equation} $$

The substitutions made into \eqref{simple_integral} create the following integral:

$$ \begin{equation}\label{simple_integral_pointtran} \int_{\arctan(a)}^{\arctan(y)} \sec^3(\theta) \, d\theta \end{equation} $$

Which is a whole different ball of wax.

[1] Scriba, Christoph J. (1963). The inverse method of tangents: A dialogue between Leibniz and Newton. Archive for History of Exact Sciences 2 (2):113-137.

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