Methods of Integration

(cod: P-70-51-3) Substitution in definite integrals: Let \(u = g(x)\) be a function with a continuous derivative on the interval \([a,b]\) and \(f\) a continuous function on an interval \(J \supset g([a,b])\). We have that \begin{equation*}\label{subst} \int_a^b f(g(x))g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du. \end{equation*} Indeed, let \(F(u)\) be a primitive for the function \(f(u)\) on the interval \(J\). By the Fundamental Theorem of Calculus, we have that \[\int_{g(a)}^{g(b)} f(u) \, du = F(g(b)) - F(g(a)).\] On the other hand, by the Chain Rule, we have that \begin{eqnarray*}\dfrac{d}{dx}F(g(x)) &=& F'(g(x))g'(x) \\ & & \\ &=& f(g(x))g'(x).\end{eqnarray*} Thus, again by the Fundamental Theorem of Calculus, we have that \[\int_a^b f(g(x))g'(x) \, dx = F(g(b)) - F(g(a)),\] which proves the statement.

Often, when performing the substitution \(u = g(x)\) in the context above, we want to replace the integral on the left-hand side with the integral on the right-hand side in the first equation. In other cases, starting from an integral \[\int_A^B f(x) \, dx\] we choose to make a variable change \begin{eqnarray*} x &=& h(\theta), \theta \in I, \\ dx &=& h'(\theta) \, d\theta, \end{eqnarray*} where we take \(h\) injective (invertible) on the interval \(I\), with \(A, B \in h(I)\). By the same argument used above, we then obtain that \[\int_A^B f(x) \, dx = \int_{a}^{b} f(h(\theta)) h'(\theta) \, d\theta,\] where \begin{eqnarray*} A = h(a) &\iff& a = h^{-1}(A) \\ & & \\ B = h(b) &\iff& b = h^{-1}(B), \end{eqnarray*} allowing us to replace the integral on the left-hand side with the one on the right-hand side. The difference in this case compared to the previous one is that, while previously the introduced variable was written as a function of the original variable (\(u = g(x)\)), here we have the opposite, that is, we write the original variable as a function of the new integration variable (\(x = h(\theta)\)).

In the questions below, consider that \(f\) is a continuous function:

(1) By making the substitution \(u = x^2 + 1\), we can verify that the integral \[\int_0^1 2x f(x^2+1) \, dx\] is equal to:

(2) By making the substitution \(x = \sin\, \theta, \theta \in [-\pi/2,\pi/2]\), we can verify that the integral \[\int_0^{1/2} f(x) \, dx\] is equal to:

(cod: P-70-71-15) Consider the following questions:

(1) Calculate \(\displaystyle \int \dfrac{1}{x \ln^2{x}}\, dx\).

(2) What can be stated about the improper integral \(\displaystyle \int_0^{+ \infty}x e^{-x^2}\, dx\)?

(3) Select an integral that has the same value as \(\displaystyle \int_0^{\frac{\pi}{2}} e^{\sin\, {x}}\, \sin(2x)\, dx\).

(4) Determine \(\displaystyle \int \tan{x}\, dx\).

(cod: P-71-62-1) Integration by Parts: This is one of the main techniques of integration, being a consequence of the product rule. Indeed, let \(f\) and \(g\) be two functions with continuous derivatives on an open interval \(I\). By the Product Rule, we have \[\dfrac{d}{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x),\] or, equivalently, \[\small{f(x)g'(x) = \dfrac{d}{dx} \left(f(x)g(x)\right) - g(x)f'(x).} \ \ \ \ \ (1)\] Taking the antiderivative on both sides, we obtain \[\small{\int f(x)g'(x)\, dx = f(x)g(x) - \int g(x)f'(x)\, dx},\] where we omit the constant of integration when integrating the first term on the right-hand side, as this constant will be introduced when solving the integral. The technique consists of replacing the integral on the left-hand side with the expression on the right-hand side. Taking \(u = f(x)\) and \(dv = g'(x)dx\) in the integral on the left-hand side and considering that \(du = f'(x)dx\) and \(v = g(x)\), we can rewrite the above formula in a more compact form: \[\int u \, dv = uv - \int v \, du.\] When calculating a definite integral over an interval \([a,b] \subset I\), we can integrate both sides of equation (1), obtaining \begin{eqnarray*} & & \int_a^b f(x)g'(x)\, dx \\ & & \\ &=& f(x)g(x)\Big{]}_a^b + \int_a^b g(x)f'(x)\, dx. \end{eqnarray*} Consider the questions below:

(1) Suppose we want to calculate the integral \[\int x e^x \, dx\] using integration by parts and, for this, we set \(u = x\) and \(dv = e^x dx\). What will be our conclusion?

(2) Suppose we want to calculate the integral \[\int_1^e \ln{x} \, dx\] using integration by parts and, for this, we choose \(u = \ln{x}\) and \(dv = dx\) (or, equivalently, \(dv/dx = 1\)). What will we obtain?

(cod: P-71-64-11) Determine the integrals below:

(1) \(\displaystyle \int x \cos{x} \, dx\)

(2) \(\displaystyle \int_0^{\frac{\pi}{2}}e^x \cos{x} \, dx\)

(3) \(\displaystyle \int_0^1 \arctan\, x \, dx\).

(cod: P-74-59-2) Integration by Partial Fractions: Consider the sum below \[\dfrac{3}{x} + \dfrac{1}{x - 2} = \dfrac{3(x-2) + x}{x^2 - x} = \dfrac{4x - 6}{x^2 - x}.\] If we wanted to integrate the rational function \(g(x) = \dfrac{4x - 6}{x^2 - x}\), it would be interesting to work from the inverse perspective and consider it as the sum of simpler fractions that appear on the left side of the first equality.

To illustrate the process, let us take the rational function \[f(x) = \dfrac{-2x+1}{x^2 - 3x + 2}.\] The denominator \(Q(x) = x^2 - 3x + 2\) has two distinct real roots, namely, \(x = 1\) and \(x = 2\). Thus, \(Q(x)\) is reducible, meaning it can be factored as the product of two lower-degree polynomials (in this case, two polynomials of degree \(1\)): \[Q(x) = x^2 - 3x + 2 = (x-1)(x-2).\] We would then like to find constants \(A\) and \(B\) such that \[\dfrac{- 2x + 1}{x^2 - 3x + 2} = \dfrac{A}{x-1} + \dfrac{B}{x-2}.\] We can multiply both sides by \(x^2 - 3x + 2 = (x-1)(x-2)\), obtaining \begin{equation}\label{igual}- 2x+ 1 = A(x-2) + B(x-1).\end{equation} Grouping the terms on the right-hand side by degree, we get \[- 2x + 1 = (A+B)x + (-2A-B).\] Comparing the coefficients of the polynomials on both sides, we must have \[\left\{\begin{array}{l} A+B = -2 \\ -2A-B = 1 \end{array}\right.\] Solving the system, we find that \(A = 1\) and \(B=-3\). In this example, we could have proceeded differently to obtain the constants \(A\) and \(B\). Indeed, since we want the equality in (\ref{igual}) to hold for any value of \(x \in \mathbb{R}\), we can substitute specific values and easily find the values of \(A\) and \(B\). More precisely, setting \(x = 1\) in (\ref{igual}), we directly find that \(A = 1\), while setting \(x=2\), we conclude that \(B = -3\). Therefore, we have that \begin{eqnarray*} & & \int \dfrac{-2x + 1}{x^2 - 3x + 2} \, dx \\ & & \\ &=& \int \left(\dfrac{1}{x-1} - \dfrac{3}{x-2} \right) \, dx \\ & & \\ &=& \ln|x-1| -3\ln|x-2| + C. \end{eqnarray*} This example illustrates the method of partial fractions. It is a method for integrating rational functions that involves expressing a function of this class (which is a ratio of polynomials) as the sum of simpler fractions (called partial fractions), which we can integrate. We then proceed to describe the method in a more systematic way.

We say that a rational function is proper if the degree of the numerator is less than the degree of the denominator. Otherwise, we say that the rational function is improper. It is always possible to write an improper rational function \(h(x) = \dfrac{P(x)}{Q(x)}\) as the sum of a polynomial and a proper rational function by applying the division algorithm. Indeed, if the quotient of the division of \(P(x)\) by \(Q(x)\) is \(A(x)\), with remainder \(R(x)\), we have \[P(x) = Q(x)A(x) + R(x),\] where the degree of \(R(x)\) is less than the degree of \(Q(x)\). Dividing the above equality by \(Q(x)\), we then obtain \[\dfrac{P(x)}{Q(x)} = A(x) + \dfrac{R(x)}{Q(x)},\] where the ratio on the right-hand side is a proper rational function.

We will describe the method considering proper rational functions, applying the above procedure if the function is improper. It is possible to demonstrate that the decomposition strategy presented below always works, in the sense that it is always possible to determine the constants and, finally, integrate all the partial fractions that arise.

From now on, let us consider a proper rational function \(f(x) = \dfrac{P(x)}{Q(x)},\) where the numerator and denominator are coprime, that is, they have no common factors. The decomposition of \(f\) as a sum of partial fractions depends on the factorization of \(Q(x)\) into irreducible factors. We have the following fact:

Every polynomial with real coefficients admits a factorization as a product of irreducible factors of degree 1 or 2 (although, in practice, it is difficult to exhibit such a factorization).

We now briefly explain the above fact. The Fundamental Theorem of Algebra guarantees that \(Q(x)\) has \(n\) roots (not necessarily distinct) in the set of complex numbers, some of which may be real. Starting from the \(n\) roots \(z_1, z_2, \dots,z_n\) of \(Q(x)\) in \(\mathbb{C}\), we first verify that \(Q(x)\) admits a decomposition (using complex numbers) in the form \begin{equation*}\label{dec1}Q(x) = c_n(x - z_1) (x-z_2)\cdots (x-z_n), \end{equation*} where \(c_n \in \mathbb{R}\) is the leading coefficient of \(Q(x)\) (note that we can consider \(c_n = 1\) after a simple algebraic manipulation in \(f(x)\)). Since \(Q(x)\) has real coefficients, it is possible to ensure that, for each non-real complex root \(z\) of \(Q(x)\), its conjugate \(\overline{z}\) will also be a root, simply by using that \[Q(\overline{z}) = \overline{Q(z)} = \overline{0} = 0.\] Note, then, that the product of the factors \((x - z)(x-\overline{z})\) gives rise to the quadratic polynomial, \[x^2 - \underbrace{(z+\overline{z})}_{\in \mathbb{R}} x + \underbrace{z\overline{z}}_{\in \mathbb{R}},\] which has real coefficients and whose roots are precisely \(z\) and \(\overline{z}\) (negative discriminant). Thus, starting from the above decomposition, we can obtain the mentioned factorization, where each real root of \(Q(x)\) gives rise to a degree 1 factor, while non-real roots appear in pairs (conjugates), giving rise to an irreducible quadratic factor.

Grouping the repeated factors (roots with multiplicity greater than 1), we can consider the factorization \[Q(x) = q_1(x)q_2(x) \cdots q_m(x),\] where a block \(q_j(x)\) can take the form \begin{equation}\label{bloco1}(x - r)^{l} \end{equation} or \begin{equation}\label{bloco2}(x^2 + bx + c)^{k}, \end{equation} with the first case associated with a real root \(r\) of multiplicity \(l\), and the second associated with a pair \(z\) and \(\overline{z}\) of conjugate complex roots, each with multiplicity \(k\).

From these blocks, we establish the decomposition of the proper rational function \(f(x) = \dfrac{P(x)}{Q(x)}\) as a sum of partial fractions. For each block of the form (\ref{bloco1}), we add \(l\) partial fractions to the decomposition: \[\dfrac{A_1}{x-r} + \dfrac{A_2}{(x-r)^2} + \cdots + \dfrac{A_l}{(x-r)^l},\] where \(A_1,A_2,\dots,A_l\) are constants to be determined. On the other hand, for each block of the form (\ref{bloco2}), we add \(k\) partial fractions to the decomposition: \[\scriptsize{\dfrac{B_1 x + C_1 }{x^2 + bx + c} + \dfrac{B_2 x + C_2 }{(x^2 + bx + c)^2} + \cdots + \dfrac{B_k x + C_k }{(x^2 + bx + c)^k},}\] where \(B_j\) and \(C_j\), with \(j = 1,\dots,k\), are constants to be determined. The decomposition into partial fractions is given by the sum of all fractions, considering all blocks.

To make it clearer, consider an example where the denominator factors as \begin{eqnarray*}Q(x) &=& (x-1)(x-1)(x^2+4) \\ & & \\ &=& (x-1)^2 (x^2+4). \end{eqnarray*} The degree \(1\) factor appears twice (\(x = 1\) is a double root), while \(x^2+4\) is an irreducible quadratic factor (whose roots are \(z = 2i\) and \(\overline{z} = - 2i\)). The decomposition into partial fractions of \(f\) would then be \[\dfrac{P(x)}{Q(x)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2 + 4}.\] Multiplying both sides by \(Q(x)\) and comparing the coefficients on the left and right sides, we can calculate the constants \(A\), \(B\), \(C\), and \(D\), and then calculate the integral of \(f\).

Select the option that does not follow the instructions given above.

(cod: P-76-83-8) The Laplace transform of a function \(f: [0,+\infty) \to \mathbb{R}\) is the function given by \[\mathcal{L}\{f(t)\}(s) = \int_0^{+\infty} e^{-st}f(t) \, dt,\] which is defined only for the values of \(s\) where the improper integral converges.

For example, consider the constant function \(f(t) = 1, t \geq 0\). In this case, for \(s>0\), we have: \begin{eqnarray*}\mathcal{L}\{f(t)\} &=& \int_0^{+ \infty} e^{-st} \, dt \\ &=& \lim\limits_{A \to + \infty} \int_0^A e^{-st} \, dt \\ &=& \lim\limits_{A \to + \infty} \left(\frac{1}{s} - \frac{e^{-sA}}{s}\right) \\ &=& \frac{1}{s}. \end{eqnarray*} Consider the following questions:

(1) Determine \(\mathcal{L}\{\sin(at)\}\), where \(a\) is a non-zero constant.

(2) Let \(f\) be a differentiable function with a continuous derivative on \([0,+\infty)\). Suppose that the transform of \(f\) is defined for \(s>0\) and that \[\lim\limits_{t \to + \infty} f(t)e^{-st} = 0,\] for all \(s>0\). In this case, it can be verified that the Laplace transform of \(f'(t)\) is also defined for \(s>0\). Select the option that presents a correct relationship between the transforms of \(f\) and \(f'\).

(cod: P-73-57-2) Trigonometric substitutions: Integrals involving the radical expressions \[ \sqrt{a^2 - x^2}, \sqrt{a^2 + x^2}, \text{ or } \sqrt{x^2-a^2},\] where \(a\) is a positive number, can often be transformed into simpler integrals through an appropriate trigonometric substitution. Indeed, consider the trigonometric identities below: \begin{eqnarray} \label{rel1} 1 - \sin^2\, \theta &=& \cos^2{\theta}, \\ \nonumber & & \\ \label{rel2} 1 + \tan^2{\theta} &=& \sec^2{\theta}, \\ \nonumber & & \\\label{rel3} \sec^2{\theta} - 1 &=& \tan^2{\theta}. \end{eqnarray} If the integral involves the radical \(\sqrt{a^2 - x^2}\), it may be convenient to use the substitution \begin{eqnarray*} x &=& a \, \sin\, \theta, \, \, \theta \in [-\pi/2,\pi/2], \\ dx &=& a \cos{\theta} \, d\theta, \end{eqnarray*} Since, by relation (\ref{rel1}), we have \begin{eqnarray*}\sqrt{a^2 - x^2} &=& \sqrt{a^2(1 - \sin^2\, {\theta})} \\ & & \\ &=& \sqrt{a^2 \cos^2{\theta}} = a \cos{\theta}. \end{eqnarray*} On the other hand, if the integral involves the radical \(\sqrt{a^2 + x^2}\), it may be convenient to use the substitution \begin{eqnarray*} x &=& a \, \tan{\theta}, \, \, \theta \in (-\pi/2,\pi/2), \\ dx &=& a \sec^2{\theta} \, d\theta, \end{eqnarray*} Since, by relation (\ref{rel2}), we have \begin{eqnarray*}\sqrt{a^2 + x^2} &=& \sqrt{a^2(1 + \tan^2\, {\theta})} \\ & & \\ &=& \sqrt{a^2 \sec^2{\theta}} = a \sec{\theta}. \end{eqnarray*} Finally, if the integral involves the radical \(\sqrt{x^2 - a^2}\), it may be convenient to use the substitution \begin{eqnarray*} x &=& a \, \sec{\theta}, \, \, \small{\theta \in (0,\pi/2) \text{ or } \theta \in (\pi,3\pi/2)}, \\ dx &=& \sec^2{\theta} \, d\theta, \end{eqnarray*} Since, by relation (\ref{rel3}), we have \begin{eqnarray*}\sqrt{x^2-a^2} &=& \sqrt{a^2(\sec^2{\theta} - 1)} \\ & & \\ &=& \sqrt{a^2 \tan^2{\theta}} = a \tan{\theta}. \end{eqnarray*} It is worth noting that these substitutions may not guarantee that the integral becomes simple, depending on the complete expression of the integrand. On the other hand, we observe that these substitutions can also be useful in cases where the root does not appear in the integral, that is, they are not exclusive to the cases addressed above, although they are more commonly used in this context.

Using an appropriate trigonometric substitution, we can ensure that the integral \[\int_{0}^{\frac{a\sqrt{2}}{2}} \dfrac{a}{\sqrt{a^2 - x^2}}\, dx\] where \(a\) is a positive number, is equal to: