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.

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

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-eq}, is all we need for IBP. The article section on it can be reviewed here: Product Rule .

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

The way the Integration By Parts formula, eq. \eqref{IBP-eq}, 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.

$$ \begin{equation}\label{simple-integral} F = \int_a^b \sqrt{1+x^2} \, dx \end{equation} $$

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-eq}, 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 transform it with a change of variable into a form with more options for identity usage.

Point Transformation

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 $[-\pi/2 + 0.071, \pi/2 - 0.071]$.

Integrand Substitution

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 tangent of angle, $\theta$. If two functions map separate domains to the same range via continuous bijections, they are compatible substitutes.

$$ \begin{equation}\label{simple-integral-point-transform} x = \tan(\theta) \end{equation} $$

Which makes the integrand a function of tangent, like so:

$$ \begin{equation}\label{simple-integral-point-transform-integrand} f=\sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = \sec(\theta) \end{equation} $$

Differential Substitution

Changing the variable has consequences for the integration beyond the integrand to the differential and endpoints. Recall that the width of the differential slice of the area under the curve in this FTC inductive figure is some width in space $x$, but $dx$ transforms as its own function into the new transformed differential slice.

$$ \begin{equation}\label{simple-integral-diff-transform} dx = g(\theta)d\theta \end{equation} $$

Where the function, $g(\theta)$, is determined by the derivative of the point transformation, $dx/d\theta$.

The differential unit, $dx$, is transformed with a derivative, which is the same for every point transformation. Here we will use the definition of tangent, the Product Rule, and the two derivatives of the cofunctions, and finally the definition of secant.

$$ \begin{equation}\label{simple-integral-diff} \frac{dx}{d\theta} = \frac{d}{d\theta}\tan(\theta)= \frac{d}{d\theta}\frac{\sin(\theta)}{\cos(\theta)} = 1 + \tan^2(\theta) = \sec^2(\theta) \end{equation} $$

With the substitution of the differential being in the following way:

$$ \begin{equation}\label{simple-integral-diff-sub} dx = \sec^2(\theta) \, d\theta \end{equation} $$

Which determines the unknown differential, \eqref{simple-integral-diff-transform}, to be, $g(\theta)=\sec^2(\theta)$.

Endpoint Substitution

The endpoints are transformed with the substitution equivalency, \eqref{simple-integral-point-transform}, itself.

$$ \begin{equation}\label{simple-integral-endpoints} \theta_0 = \arctan(a)\quad , \quad \theta_1 = \arctan(b) \end{equation} $$

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

$$ \begin{equation}\label{simple-integral-pointtran} F=\int_{\theta_0}^{\theta_1} \sec^3(\theta) \, d\theta \end{equation} $$

Which is a whole different ball of wax, because it's got more trigonometric combinations as options with the cubed form of the secant than the original integral form.

Applying IBP

Considering the derivative of tangent, which we calculated in \eqref{simple-integral-diff}, we see that the integrand of the (point) transformed integral now has the form of a product of a derivative and another term, which is the product form of IBP.

$$ f\, d\theta=(\sec(\theta))(\sec^2(\theta)\, d\theta ) = u\, dv $$

Where we've grouped the derivative together with the differential unit and assigned it to the IBP's $dv$ term, and the other remaining factor is assigned to the IBP's $u$ term. So we calculate the derivative of $u$ to get $du$, and the antiderivative of $dv$ to get $v$.

$$ u=\sec(\theta) \quad , \quad \frac{du}{d\theta}=\tan(\theta) \sec(\theta) $$

Calculating the derivative of the $u$ term gives us the $du$ term (always the easy part). As the $dv$ term is a known integrand, we can immediately write its antiderivative term, $v$.

$$ \frac{dv}{d\theta} = \sec^2(\theta) \quad , \quad v = \tan(\theta) $$

We plug these terms into \eqref{IBP-eq} and rearrange the three terms of it in order to transform the original integral into a definite integral term and a new integral.

$$ \begin{equation}\label{simple-integral-IBP1} \int_{\theta_0}^{\theta_1} \sec^3(\theta) \, d\theta = \left. \tan(\theta) \sec(\theta) \right|_{\theta_0}^{\theta_1} - \int_{\theta_0}^{\theta_1} \tan^2(\theta)\sec(\theta)\, d\theta \end{equation} $$

The remaining problem is solving the integral on the RHS. The new integrand is $f=\tan^2(\theta)\sec(\theta)$. You can't immediately apply IBP twice in a row, because you just undo the previous step. To solve this integral we first use the trigonometric identity of tangent squared, in \eqref{simple-integral-diff}.

$$ \int_{\theta_0}^{\theta_1} \tan^2(\theta)\sec(\theta), d\theta = \int_{\theta_0}^{\theta_1} (\sec^3(\theta)-\sec(\theta)), d\theta $$

Which gives us another term of $\sec^3(\theta)$ on the RHS, but since they have different signs they don't cancel.

Applying Trigonometric Identities

Now the second appearance of the $\sec^3$ term means we have to substitute this and algebraically rearrange \eqref{simple-integral-IBP1}, to isolate an integral of secant as the next step. Since a single term in the integrand is hard to work with (unless it's trivial, like a polynomial), so we'll again transform it using trigonmetric identities and algebraic rules.

$$ \begin{equation}\label{simple-integral-tan2sub} 2F = 2\int_{\theta_0}^{\theta_1} \sec^3(\theta) \, d\theta = \left. \tan(\theta) \sec(\theta) \right|_{\theta_0}^{\theta_1} - \int_{\theta_0}^{\theta_1} \sec(\theta)\, d\theta \end{equation} $$

The integrand of a single secant term is not amenable to IBP since it doesn't have the product form, but we can use the Pythagorean Identity and some algebra to get an integral of the hyperbola, which we know to be the natural logarithm.

$$ \begin{equation}\label{simple-integral-secant-algebra} \int_{\theta_0}^{\theta_1} \frac{\cos(\theta)}{\cos^2(\theta)}\, d\theta = \int_{\theta_0}^{\theta_1} \frac{\cos(\theta)}{1-\sin^2(\theta)}\, d\theta \end{equation} $$

Since cosine is the derivative of sine, this last integral form is promising because if we can get the denominator of the RHS of \eqref{simple-integral-secant-algebra} into an easier form of sine, the numerator provides the complimentary differential.

Difference of Two Squares

Whenever you see a term in the form of the difference of two squares, such as $(1-x^2)$, this is amenable to the algebraic rule of the Difference of Two Squares:

$$ \begin{equation}\label{difference-of-2-squares} (A^2 - B^2) = (A-B)(A+B) \end{equation} $$

Or for this case, we have:

$$ (1-\sin^2(\theta))=(1-\sin(\theta))(1+\sin(\theta)) $$

When the difference of two squares is in the denominator of \eqref{simple-integral-secant-algebra}, there is an algebraic simplification we can use, called Partial Fraction Decomposition.

Partial Fraction Decomposition

In general, a fraction with a product of two factors in the denominator can be decomposed into two fractions.

$$ \begin{equation}\label{partial-fraction-decomposition-eq} \frac{1}{AB} = \frac{\alpha}{A} + \frac{\beta}{B} \end{equation} $$

Where, $\alpha B + \beta A = 1$. For $A+B=2$, as in this case, we have $\alpha=\beta=\frac{1}{2}$. Substituting this into the RHS of \eqref{simple-integral-secant-algebra}:

$$ \begin{equation}\label{simple-integral-secant-diff2squares} \int_{\theta_0}^{\theta_1} \frac{\cos(\theta)}{1-\sin^2(\theta)}\, d\theta = \int_{\theta_0}^{\theta_1} \left(\frac{\cos(\theta)}{2(1-\sin(\theta))}+\frac{\cos(\theta)}{2(1+\sin(\theta))}\right)\, d\theta \end{equation} $$

Which you can recognize as two hyperbolic integrands ready to be integrated immediately.

Hyperbolic Integrand

Since the numerators in the RHS of \eqref{simple-integral-secant-diff2squares}, $du=\cos(\theta)\, d\theta$, are the differentials of the denominators, with an additional negative sign on the one with negative sine, we know the derivative of logarithm, combined with the Fundamental Theorem of Calculus, we know the antiderivative of the hyperbola is the logarithm.

$$ \begin{equation}\label{simple-integral-secant-PFD} \int_{\theta_0}^{\theta_1} \sec(\theta)\, d\theta =\int_{\theta_0}^{\theta_1} \left(\frac{\cos(\theta)}{2(1-\sin(\theta))}+\frac{\cos(\theta)}{2(1+\sin(\theta))}\right)\, d\theta =\frac{1}{2}\left.-\ln(1-\sin(\theta)) + \ln(1+\sin(\theta))\right|_{\theta_0}^{\theta_1} =\frac{1}{2}\left.\ln\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)\right|_{\theta_0}^{\theta_1} =\frac{1}{2}\left.\ln\left(\frac{(1+\sin(\theta))^2}{1-\sin^2(\theta)}\right)\right|_{\theta_0}^{\theta_1} =\frac{1}{2}\left.\ln\left(\frac{(1+\sin(\theta))^2}{\cos^2(\theta)}\right)\right|_{\theta_0}^{\theta_1} =\left.\ln\left(\frac{(1+\sin(\theta))}{\cos(\theta)}\right)\right|_{\theta_0}^{\theta_1} =\left.\ln(\sec(\theta)+\tan(\theta))\right|_{\theta_0}^{\theta_1} \end{equation} $$

The integral is almost complete after plugging \eqref{simple-integral-secant-PFD} into \eqref{simple-integral-tan2sub}, save for the reversal of the point transformation.

Reversal of the Point Transformation

The penultimate form of this integral is in \eqref{simple-integral-penultimate}.

$$ \begin{equation}\label{simple-integral-penultimate} F = \frac{1}{2}\left. \tan(\theta) \sec(\theta) \right|_{\theta_0}^{\theta_1} + \frac{1}{2}\left.\ln(\sec(\theta)+\tan(\theta))\right|_{\theta_0}^{\theta_1} \end{equation} $$

We reverse the point transformation by converting the angular endpoints and trigonometric functions back to the originals.

$$ \tan(\theta) = x $$

$$ \sec(\theta) = \sqrt{1+\tan^2(\theta)} = \sqrt{1+x^2} $$

$$ x_0 = \tan(\theta_0) = \tan(\arctan(a))=a \quad , \quad x_1= \tan(\theta_1) = \tan(\arctan(b))=b $$

$$ \begin{equation}\label{simple-integral-ultimate} F = \frac{1}{2}\left. x\sqrt{1+x^2} \right|_{a}^{b} + \frac{1}{2}\left.\ln(x + \sqrt{1+x^2})\right|_{a}^{b} \end{equation} $$

The integral \eqref{simple-integral} is completed with \eqref{simple-integral-ultimate}. Note there are always more than one way to solve an integral, or any problem for that matter.

Graph of the Integral

Figure 2. Graph of the integral of the simple function on the interval, $[0,10]$.

The normalized binomial of power one half is approximated by the identity function, so over the interval, of Figure 2 (above), the integral acquires a value in the neighborhood of $F\approx\frac{1}{2}10^2=50$, which aligns with the values of $F$ as the dashed line, for sanity check.

Figure 3. Graph of the integral of the simple function on the subinterval, $[0,3]$.

[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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