Integration by substitution

Introduction

In Section, we learned the key role that antiderivatives play in the process of evaluating definite integrals exactly. The Fundamental Theorem of Calculus tells us that if \(F\) is any antiderivative of \(f\), then \[\begin{aligned}\end{aligned}\].

Furthermore, we realized that each elementary derivative rule developed in Chapter leads to a corresponding elementary antiderivative, as summarized in Table. Thus, if we wish to evaluate an integral such as \[\begin{aligned}\end{aligned}\], it is straightforward to do so, since we can easily antidifferentiate \(f(x) = x^3 - \sqrt{x} + 5^x\). Because one antiderivative of \(f\) is \(F(x) = \frac{1}{4}x^4 - \frac{2}{3}x^{3/2} + \frac{1}{\ln(5)}5^x\), the Fundamental Theorem of Calculus tells us that \[\begin{aligned}\int_0^1 \left(x^3 - \sqrt{x} + 5^x\right) \,dx \amp= \left. \frac{1}{4}x^4 - \frac{2}{3}x^{3/2} + \frac{1}{\ln(5)}5^x\right|_0^1 \\ \amp= \left( \frac{1}{4}(1)^4 - \frac{2}{3}(1)^{3/2} + \frac{1}{\ln(5)}5^1 \right) - \left( 0 - 0 + \frac{1}{\ln(5)}5^0 \right) \\ \amp= -\frac{5}{12} + \frac{4}{\ln(5)}\end{aligned}\].

We see that we have a natural interest in being able to find such algebraic antiderivatives. We emphasize algebraic antiderivatives, as opposed to any antiderivative, since we know by the Second Fundamental Theorem of Calculus that \(G(x) = \int_a^x f(t) \, dt\) is indeed an antiderivative of the given function \(f\), but one that still involves a definite integral. Our goal in this section is to undo the process of differentiation to find an algebraic antiderivative for a given function.

Exploration

Exploration

Reversing the Chain Rule: First Steps

Whenever \(f\) is a familiar function whose antiderivative is known and \(u(x)\) is a linear function, it is straightforward to antidifferentiate a function of the form \[\begin{aligned}\end{aligned}\].

Example

Determine the general antiderivative of \[\begin{aligned}\end{aligned}\]. Check the result by differentiating.

Solution

For this composite function, the outer function \(f\) is \(f(u) = u^6\), while the inner function is \(u(x) = 5x - 3\). Since the antiderivative of \(f\) is \(F(u) = \frac{1}{7}u^7+C\), we see that the antiderivative of \(h\) is \[\begin{aligned}\end{aligned}\].

The inclusion of the constant \(\frac{1}{5}\) is essential precisely because the derivative of the inner function is \(u'(x) = 5\). Indeed, if we now compute \(H'(x)\), we find by the Chain Rule (and Constant Multiple Rule) that \[\begin{aligned}\end{aligned}\], and thus \(H\) is indeed the general antiderivative of \(h\).

Hence, in the special case where the outer function is familiar and the inner function is linear, we can antidifferentiate composite functions according to the following rule.

If \(h(x) = f(ax + b)\) and \(F\) is a known algebraic antiderivative of \(f\), then the general antiderivative of \(h\) is given by \[\begin{aligned}\end{aligned}\].

It is useful to have shorthand notation that indicates the instruction to find an antiderivative. Thus, in a similar way to how the notation \[\begin{aligned}\end{aligned}\] represents the derivative of \(f(x)\) with respect to \(x\), we use the notation of the indefinite integral, \[\begin{aligned}\end{aligned}\] to represent the general antiderivative of \(f\) with respect to \(x\). Returning to the earlier example with \(h(x) = (5x-3)^6\), we can rephrase the relationship between \(h\) and its antiderivative \(H\) through the notation \[\begin{aligned}\end{aligned}\].

When we find an antiderivative, we will often say that we evaluate an indefinite integral. Just as the notation \(\frac{d}{dx} [ \Box ]\) means find the derivative with respect to \(x\) of \(\Box\), the notation \(\int \Box \, dx\) means find a function of \(x\) whose derivative is \(\Box\).

Condensed: the full section is in Boelkins, Active Calculus.

Reversing the Chain Rule: u-substitution

A natural question arises from our recent work: what happens when the inner function is not linear? For example, can we find antiderivatives of such functions as \[\begin{aligned}\end{aligned}\]

It is important to remember that differentiation and antidifferentiation are almost inverse processes (that they are not is due to the \(+C\) that arises when antidifferentiating). This almost-inverse relationship enables us to take any known derivative rule and rewrite it as a corresponding rule for an indefinite integral. For example, since \[\begin{aligned}\end{aligned}\], we can equivalently write \[\begin{aligned}\end{aligned}\].

Recall that the Chain Rule states that \[\begin{aligned}\end{aligned}\].

Restating this relationship in terms of an indefinite integral, \[\begin{aligned}\end{aligned}\].

Equation tells us that if we can view a given function as \(f'(g(x)) g'(x)\) for some appropriate choices of \(f\) and \(g\), then we can antidifferentiate the function by reversing the Chain Rule. Note that both \(g(x)\) and \(g'(x)\) appear in the form of \(f'(g(x)) g'(x)\); we will sometimes say that we seek to identify a function-derivative pair (\(g(x)\) and \(g'(x)\)) when trying to apply the rule in Equation.

If we can identify a function-derivative pair, we will introduce a new variable \(u\) to represent the function \(g(x)\). With \(u = g(x)\), it follows in Leibniz notation that \(\frac{du}{dx} = g'(x)\), so that in terms of differentials If we recall from the definition of the derivative that \(\frac{du}{dx} \approx \frac{\Delta{u}}{\Delta{x}}\) and use the fact that \(\frac{du}{dx} = g'(x)\), then we see that \(g'(x) \approx \frac{\Delta{u}}{\Delta{x}}\). Solving for \(\Delta u\), \(\Delta u \approx g'(x) \Delta x\). It is this last relationship that, when expressed in differential notation enables us to write \(du = g'(x) \, dx\) in the change of variable formula. , \(du = g'(x)\, dx\). Now converting the indefinite integral to a new one in terms of \(u\), we have \[\begin{aligned}\end{aligned}\].

Provided that \(f'\) is an elementary function whose antiderivative is known, we can easily evaluate the indefinite integral in \(u\), and then go on to determine the desired overall antiderivative of \(f'(g(x)) g'(x)\). We call this process \(u\)-substitution, and summarize the rule as follows:

With the substitution \(u = g(x)\), \[\begin{aligned}\end{aligned}\].

Condensed: the full section is in Boelkins, Active Calculus.

Evaluating Definite Integrals via u-substitution

We have introduced \(u\)-substitution as a means to evaluate indefinite integrals of functions that can be written, up to a constant multiple, in the form \(f(g(x))g'(x)\). This same technique can be used to evaluate definite integrals involving such functions, though we need to be careful with the corresponding limits of integration. Consider, for instance, the definite integral \[\begin{aligned}\end{aligned}\].

Whenever we write a definite integral, it is implicit that the limits of integration correspond to the variable of integration. To be more explicit, observe that \[\begin{aligned}\end{aligned}\].

When we execute a \(u\)-substitution, we change the variable of integration; it is essential to note that this also changes the limits of integration. For instance, with the substitution \(u = x^2\) and \(du = 2x \, dx\), it also follows that when \(x = 2\), \(u = 2^2 = 4\), and when \(x = 5\), \(u = 5^2 = 25\). Thus, under the change of variables of \(u\)-substitution, we now have \[\begin{aligned}\int_{x=2}^{x=5} xe^{x^2} \, dx \amp= \int_{u=4}^{u=25} e^{u} \cdot \frac{1}{2} \, du \\ \amp= \left. \frac{1}{2}e^u \right|_{u=4}^{u=25} \\ \amp= \frac{1}{2}e^{25} - \frac{1}{2}e^4\end{aligned}\].

Alternatively, we could consider the related indefinite integral \(\int xe^{x^2} \, dx\), find the antiderivative \(\frac{1}{2}e^{x^2}\) through \(u\)-substitution, and then evaluate the original definite integral. With that method, we'd have \[\begin{aligned}\int_{2}^{5} xe^{x^2} \, dx \amp= \left. \frac{1}{2}e^{x^2} \right|_{2}^{5} \\ \amp= \frac{1}{2}e^{25} - \frac{1}{2}e^4\end{aligned}\], which is, of course, the same result.

Summary

• To find algebraic formulas for antiderivatives of more complicated algebraic functions, we need to think carefully about how we can reverse known differentiation rules. To that end, it is essential that we understand and recall known derivatives of basic functions, as well as the standard derivative rules.

• The indefinite integral provides notation for antiderivatives. When we write \(\int f(x) \, dx\), we mean the general antiderivative of \(f\). In particular, if we have functions \(f\) and \(F\) such that \(F' = f\), the following two statements say the exact thing: \[\begin{aligned}\end{aligned}\]. That is, \(f\) is the derivative of \(F\), and \(F\) is an antiderivative of \(f\).

• The technique of \(u\)-substitution helps us to evaluate indefinite integrals of the form \(\int f(g(x))g'(x) \, dx\) through the substitutions \(u = g(x)\) and \(du = g'(x) \, dx\), so that \[\begin{aligned}\end{aligned}\]. A key part of choosing the expression in \(x\) to be represented by \(u\) is the identification of a function-derivative pair. To do so, we often look for an inner function \(g(x)\) that is part of a composite function, while investigating whether \(g'(x)\) (or a constant multiple of \(g'(x)\)) is present as a multiplying factor of the integrand.

Integration by substitution

In calculus, integration by substitution, also known as u-substitution, reverse chain rule or change of variables, is a method for evaluating integrals and antiderivatives. It is the counterpart to the chain rule for differentiation, and can loosely be thought of as using the chain rule "backwards." This involves differential forms.

Introduction (indefinite integrals)

Before stating the result rigorously, consider a simple case using indefinite integrals.

Compute \(\int(2x^3+1)^7(x^2)\,dx.\)

Set \(u=2x^3+1.\) This means \(\frac{du}{dx}=6x^2,\) or as a differential form, \(du=6x^2\,dx.\) Now consider \(\int u^7du.\):

\(\begin{aligned} \int(2x^3 +1)^7(x^2)\,dx \\ = \frac{1}{6}\int\underbrace{(2x^3+1)^{7}}_{u^{7}}\underbrace{(6x^2)\,dx}_{du} \\ = \frac{1}{6}\int u^{7}\,du = \frac{1}{6}\left(\frac{1}{8}u^{8}\right)+C \\ = \frac{1}{48}(2x^3+1)^{8}+C, \end{aligned}\)

where \(C\) is an arbitrary constant of integration.

This procedure is frequently used, but not all integrals are of a form that permits its use. In any event, the result should be verified by differentiating and comparing to the original integrand. \[\begin{aligned} \frac{d}{dx}\left[\frac{1}{48}(2x^3+1)^{8}+C\right] = \frac{1}{6}(2x^3+1)^{7}(6x^2) = (2x^3+1)^7(x^2). \end{aligned}\] For definite integrals, the limits of integration must also be adjusted, but the procedure is mostly the same.

Statement for definite integrals

Let \(g:[a,b]\to I\) be a differentiable function with a continuous derivative, where \(I \subset \mathbb{R}\) is an interval. Suppose that \(f:I\to\mathbb{R}\) is a continuous function. Then: \[\int_a^b f(g(x))\cdot g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\ du.\]

In Leibniz notation, the substitution \(u=g(x)\) yields: \[\frac{du}{dx} = g'(x).\] Working heuristically with infinitesimals yields the equation \[du = g'(x)\,dx,\] which suggests the substitution formula above. (This equation may be put on a rigorous foundation by interpreting it as a statement about differential forms.) One may view the method of integration by substitution as a partial justification of Leibniz's notation for integrals and derivatives.

The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be read from left to right or from right to left in order to simplify a given integral. When used in the former manner, it is sometimes known as u-substitution or w-substitution in which a new variable is defined to be a function of the original variable found inside the composite function multiplied by the derivative of the inner function. The latter manner is commonly used in trigonometric substitution, replacing the original variable with a trigonometric function of a new variable and the original differential with the differential of the trigonometric function.

Proof

Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let \(f\) and \(g\) be two functions satisfying the above hypothesis that \(f\) is continuous on \(I\) and \(g'\) is integrable on the closed interval \([a,b]\). Then the function \(f(g(x))\cdot g'(x)\) is also integrable on \([a,b]\). Hence the integrals \[\int_a^b f(g(x))\cdot g'(x)\ dx\] and \[\int_{g(a)}^{g(b)} f(u)\ du\] in fact exist, and it remains to show that they are equal.

Since \(f\) is continuous, it has an antiderivative \(F\). The composite function \(F \circ g\) is then defined. Since \(g\) is differentiable, combining the chain rule and the definition of an antiderivative gives: \[(F \circ g)'(x) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x).\]

Applying the fundamental theorem of calculus twice gives: \[\begin{aligned} \int_a^b f(g(x)) \cdot g'(x)\ dx &= \int_a^b (F \circ g)'(x)\ dx \\ &= (F \circ g)(b) - (F \circ g)(a) \\ &= F(g(b)) - F(g(a)) \\ &= \int_{g(a)}^{g(b)} F'(u)\, du = \int_{g(a)}^{g(b)} f(u)\, du, \end{aligned}\] which is the substitution rule.

Examples: Antiderivatives (indefinite integrals)

Substitution can be used to determine antiderivatives. One chooses a relation between \(x\) and \(u,\) determines the corresponding relation between \(dx\) and \(du\) by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between \(x\) and \(u\) is then undone.

Examples: Definite integrals

When evaluating definite integrals by substitution, one may calculate the antiderivative fully first, then apply the boundary conditions. In that case, there is no need to transform the boundary terms. Alternatively, one may fully evaluate the indefinite integral (see above) first then apply the boundary conditions. This becomes especially handy when multiple substitutions are used.

Substitution for multiple variables

One may also use substitution when integrating functions of several variables.

Here, the substitution function (v₁,...,v_n) = φ(u₁, ..., u_n) needs to be injective and continuously differentiable, and the differentials transform as: \[dv_1 \cdots dv_n = \left|\det(D\varphi)(u_1, \ldots, u_n)\right| \, du_1 \cdots du_n,\] where det(Dφ)(u₁, ..., u_n) denotes the determinant of the Jacobian matrix of partial derivatives of φ at the point (u₁, ..., u_n). This formula expresses the fact that the absolute value of the determinant of a matrix equals the volume of the parallelotope spanned by its columns or rows.

More precisely, the change of variables formula is stated in the next theorem:

Theorem, Let U be an open set in R and φ : U → R an injective differentiable function with continuous partial derivatives, the Jacobian of which is nonzero for every x in U. Then for any real-valued, compactly supported, continuous function f, with support contained in φ(U): \[\int_{\varphi(U)} f(\mathbf{v})\, d\mathbf{v} = \int_U f(\varphi(\mathbf{u})) \left|\det(D\varphi)(\mathbf{u})\right|\,d\mathbf{u}.\]

The conditions on the theorem can be weakened in various ways. First, the requirement that φ be continuously differentiable can be replaced by the weaker assumption that φ be merely differentiable and have a continuous inverse. This is guaranteed to hold if φ is continuously differentiable by the inverse function theorem. Alternatively, the requirement that det(Dφ) ≠ 0 can be eliminated by applying Sard's theorem.

For Lebesgue measurable functions, the theorem can be stated in the following form:

Theorem, Let U be a measurable subset of R and φ : U → R an injective function, and suppose for every x in U there exists φ′(x) in R such that φ(y) = φ(x) + φ′(x)(y − x) + o(‖y − x‖) as y → x (here o is little-o notation). Then φ(U) is measurable, and for any real-valued function f defined on φ(U): \[\int_{\varphi(U)} f(v)\, dv = \int_U f(\varphi(u)) \left|\det \varphi'(u)\right| \,du\] in the sense that if either integral exists (including the possibility of being properly infinite), then so does the other one, and they have the same value.

Another very general version in measure theory is the following:

Theorem, Let X be a locally compact Hausdorff space equipped with a finite Radon measure μ, and let Y be a σ-compact Hausdorff space with a σ-finite Radon measure ρ. Let φ : X → Y be an absolutely continuous function (where the latter means that ρ(φ(E)) = 0 whenever μ(E) = 0). Then there exists a real-valued Borel measurable function w on X such that for every Lebesgue integrable function f : Y → R, the function (f ∘ φ) ⋅ w is Lebesgue integrable on X, and \[\int_Y f(y)\,d\rho(y) = \int_X (f\circ \varphi)(x)\,w(x)\,d\mu(x).\] Furthermore, it is possible to write \[w(x) = (g\circ \varphi)(x)\] for some Borel measurable function g on Y.

Theorem, Let U be an open subset of R and φ : U → R be a bi-Lipschitz mapping. Let f : φ(U) → R be measurable. Then \[\int_{\varphi(U)} f(x)\,dx = \int_U (f\circ \varphi)(x) \left|\det D\varphi(x)\right|\,dx\] in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.

Condensed: the full section is in Wikipedia.

Application in probability

Substitution can be used to answer the following important question in probability: given a random variable X with probability density p_X and another random variable Y such that Y= ϕ(X) for injective (one-to-one) ϕ, what is the probability density for Y?

It is easiest to answer this question by first answering a slightly different question: what is the probability that Y takes a value in some particular subset S? Denote this probability P(Y ∈ S). Of course, if Y has probability density p_Y, then the answer is: \[P(Y \in S) = \int_S p_Y(y)\,dy,\] but this is not really useful because we do not know p_Y; it is what we are trying to find. We can make progress by considering the problem in the variable X. Y takes a value in S whenever X takes a value in \(\phi^{-1}(S),\) so: \[P(Y \in S) = P(X \in \phi^{-1}(S)) = \int_{\phi^{-1}(S)} p_X(x)\,dx.\]

Changing from variable x to y gives: \[P(Y \in S) = \int_{\phi^{-1}(S)} p_X(x)\,dx = \int_S p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|\,dy.\] Combining this with our first equation gives: \[\int_S p_Y(y)\,dy = \int_S p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|\,dy,\] so: \[p_Y(y) = p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|.\]

In the case where X and Y depend on several uncorrelated variables (i.e., \(p_X=p_X(x_1, \ldots, x_n)\) and \(y=\phi(x)\)), \(p_Y\)can be found by substitution in several variables discussed above. The result is: \[p_Y(y) = p_X(\phi^{-1}(y)) \left|\det D\phi ^{-1}(y) \right|.\]

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1. \[\begin{aligned}\end{aligned}\] Drag the cards from the left column into the correct labels in the right column.

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2. \[\begin{aligned}\end{aligned}\] Drag the cards from the left column into the correct labels in the right column.

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3. \[\begin{aligned}\end{aligned}\] Drag the cards from the left column into the correct labels in the right column.

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4. \[\begin{aligned}\end{aligned}\] Drag the cards from the left column into the correct labels in the right column.

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5. This problem centers on finding antiderivatives for the basic trigonometric functions other than \(\sin(x)\) and \(\cos(x)\).

   1. Consider the indefinite integral \(\int \tan(x) \, dx\). By rewriting the integrand as \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) and identifying an appropriate function-derivative pair, make a \(u\)-substitution and hence evaluate \(\int \tan(x) \, dx\).

   2. In a similar way, evaluate \(\int \cot(x) \, dx\).

   3. Consider the indefinite integral \[\begin{aligned}\end{aligned}\]. Evaluate this integral using the substitution \(u = \sec(x) + \tan(x)\).

   4. Simplify the integrand in (c) by factoring the numerator. What is a far simpler way to write the integrand?

   5. Combine your work in (c) and (d) to determine \(\int \sec(x) \, dx\).

   6. Using (c)-(e) as a guide, evaluate \(\int \csc(x) \, dx\).

   Umsebenzi Icebo

   Bonisa impendulo

   1. We let \(u = \cos(x)\), so \(du = -\sin(x) \ dx\). This change of variable transforms the integral into one we can integrate as follows: \[\begin{aligned}\int \tan(x) \, dx &= \int \frac{\sin(x)}{\cos(x)} \ dx \\ &= -\int \frac{1}{u} \ du \\ &= -\ln(|u|) + C \\ &= -\ln(|\cos(x)| + C \\ &= \ln\left(|\cos(x)|^{-1}\right) + C \\ &= \ln\left(|\sec(x)|\right) + C\end{aligned}\]. Note that we used the property of the logarithm that \(p \ln (z) = \ln(z^p)\) as well as the fact that \(\frac{1}{\cos(x)} = \sec(x)\) in order to simplify the result.

   2. We let \(u = \sin(x)\), so \(du = \cos(x) \ dx\). This change of variable transforms the integral into one we can integrate as follows: \[\begin{aligned}\int \cot(x) \, dx &= \int \frac{\cos(x)}{\sin(x)} \ dx \\ &= \int \frac{1}{u} \ du \\ &= \ln(|u|) + C \\ &= \ln(|\sin(x)| + C \\ &= -\ln\left(|\sin(x)|^{-1}\right) + C \\ &= -\ln\left(|\csc(x)|\right) + C\end{aligned}\]. Note that here we used the fact that \(\ln(z) = (-1)(-1)\ln(z) = (-1) \ln(z^{-1})\) in order to simplify the result.

   3. We let \(u = \sec(x) + \tan(x)\), so it follows that \(du = \left(\sec(x)\tan(x)+\sec^2(x)\right) \ dx\). This change of variable shows that the basic structure of the integrand is simple and can be integrated as follows: \[\begin{aligned}\int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx &= \int \frac{1}{u} \ du \\ &= \ln(|u|) + C \\ &= \ln\left(|\sec(x) + \tan(x)|\right) + C\end{aligned}\].

   4. The numerator of \(\frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)}\) can be rewritten by removing the common factor of \(\sec(x)\) to obtain \[\begin{aligned}\end{aligned}\]. The integrand can then be simplified to \[\begin{aligned}\end{aligned}\].

   5. Using the results of (c) and (d), we see that \[\begin{aligned}\end{aligned}\].

   6. Following (c)-(e) we can evaluate \(int \csc(x) \, dx\) as follows. First, rewrite \(\csc(x)\) by multiplying by \(1 = \frac{\csc(x) + \cot(x)}{\csc(x) + \cot(x)}\), so that \[\begin{aligned}\end{aligned}\]. Now we evaluate the last integral in the preceding equation with the substitution \(u = \csc(x) + \cot(x)\). Then \(du = -\left(\csc(x)\cot(x)+\csc^2(x)\right) \ dx\). This change of variable transforms the integral so that we can integrate: \[\begin{aligned}\int \frac{\csc^2(x) + \csc(x) \cot(x)}{\csc(x) + \cot(x)} \, dx &= - \int \frac{1}{u} \ dx \\ &= -\ln(|u|) + C \\ &= -\ln\left(|\csc(x) + \cot(x)|\right) + C\end{aligned}\]. Connecting the equalities we have established in our work, we have shown that \[\begin{aligned}\end{aligned}\].

6. Consider the indefinite integral \(\int x \sqrt{x-1} \, dx\).

   1. At first glance, this integrand may not seem suited to substitution due to the presence of \(x\) in separate locations in the integrand. Nonetheless, using the composite function \(\sqrt{x-1}\) as a guide, let \(u = x-1\). Determine expressions for both \(x\) and \(dx\) in terms of \(u\).

   2. Convert the given integral in \(x\) to a new integral in \(u\).

   3. Evaluate the integral in (b) by noting that \(\sqrt{u} = u^{1/2}\) and observing that it is now possible to rewrite the integrand in \(u\) by expanding through multiplication.

   4. Evaluate each of the integrals \(\int x^2 \sqrt{x-1} \, dx\) and \(\int x \sqrt{x^2 - 1} \, dx\). Write a paragraph to discuss the similarities among the three indefinite integrals in this problem and the role of substitution and algebraic rearrangement in each.

   Umsebenzi Icebo

   Bonisa impendulo

   1. Letting \(u = x-1\), we find that \(x = u+1\) and thus \(dx = du\).

   2. Using the change of variables in (a), \[\begin{aligned}\end{aligned}\].

   3. Using the fact that \(\sqrt{u} = u^{1/2}\) and distributing, we can evaluate the integral as follows: \[\begin{aligned}\int x \sqrt{x-1} \, dx &= \int (u+1) u^{1/2} \, du \\ &= \int \left(u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\].

   4. First, for \(\int x^2 \sqrt{x-1} \, dx\), we let \(u = x-1\) so that \(x = u+1\) and \(dx = du\). From there, our work is very similar to what we did in (b): \[\begin{aligned}\int x^2 \sqrt{x-1} \, dx &= \int (u+1)^2 u^{1/2} \, du \\ &= \int \left(u^2 + 2u + 1 \right) u^{1/2} \, du \\ &= \int \left(u^{5/2} + 2u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{7} u^{7/2} + \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\].

      For \(\int x \sqrt{x^2 - 1} \, dx\), despite the fact that the integral looks remarkably similar, the situation is quite different. We observe that with \(u = x^2 - 1\), it follows \(du = 2x \, dx\), so we have a standard function-derivative pair. Noting that \(x \, dx = \frac{1}{2} \, du\), it follows that \[\begin{aligned}\int x \sqrt{x^2 - 1} \, dx &= \int \sqrt{u} \cdot \frac{1}{2} \, du \\ &= \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C \\ &= \frac{1}{3} u^{3/2} + C \\ &= \frac{1}{3} (x^2-1)^{3/2} + C\end{aligned}\].

      In the situation where we have \((x-a)\) under a square root and a polynomial outside the square root, the substitution \(u = x-a\) will allow us to convert the integrand into an expression we can integrate by changing the form of the radical from \(\sqrt{x-a}\) to \(\sqrt{u}\). From there we are able to distribute and integrate an expression that is essentially a sum of (fractional) powers of \(u\). In the situation where a polynomial of degree higher than \(1\) (such as \(x^2 - 1\)) is under the radical, we likely have to be fortunate enough to have the derivative of said polynomial (up to a constant) outside the radical in order to be able to integrate, as we were in the last example.

7. Consider the indefinite integral \(\int \sin^3(x) \, dx\).

   1. Explain why the substitution \(u = \sin(x)\) will not work to help evaluate the given integral.

   2. Recall the Fundamental Trigonometric Identity, which states that \(\sin^2(x) + \cos^2(x) = 1\). By observing that \(\sin^3(x) = \sin(x) \cdot \sin^2(x)\), use the Fundamental Trigonometric Identity to rewrite the integrand as the product of \(\sin(x)\) with another function.

   3. Explain why the substitution \(u = \cos(x)\) now provides a possible way to evaluate the integral in (b).

   4. Use your work in (a)-(c) to evaluate the indefinite integral \(\int \sin^3(x) \, dx\).

   5. Use a similar approach to evaluate \(\int \cos^3(x) \, dx\).

   Umsebenzi Icebo

   Bonisa impendulo

   1. If we let \(u = \sin(x)\), then \(du = \cos(x) \, dx\). Since there is no cosine function present in the integrand, we don't have the needed function-derivative pair present in order to transform the integral into one we can evaluate.

   2. Using the Fundamental Trigonometric Identity, we observe that \(\sin^2(x) = 1-\cos^2(x)\), and thus \[\begin{aligned}\end{aligned}\].

   3. By our work in (b), we see that \(\int \sin^3(x) \, dx = \int \sin(x) (1-\cos^2(x)) \, dx\), and in this integral we have the function-derivative pair that corresponds to \(u = \cos(x)\) and \(du = -\sin(x) \, dx\).

   4. Using this substitution and evaluating the resulting integral, \[\begin{aligned}\int \sin^3(x) \, dx &= \int \sin(x) (1-\cos^2(x)) \, dx \\ &= \int (-1)(1-u^2) \, du \\ &= \int (u^2-1) \, du \\ &= \frac{1}{3}u^3 - u + C \\ &= \frac{1}{3}\cos^3(x) - \cos(x) + C\end{aligned}\].

   5. Here we observe that \(\cos^3(x) = \cos(x) \cdot \cos^2(x) = \cos(x) (1-\sin^2(x))\), so letting \(u = \sin(x)\) and thus \(du = \cos(x) \, dx\), we have \[\begin{aligned}\int \cos^3(x) \, dx &= \int \cos(x) (1-\sin^2(x)) \, dx \\ &= \int (1-u^2) \, du \\ &= u - \frac{1}{3}u^3 + C \\ &= \sin(x) - \frac{1}{3}\sin^3(x) + C\end{aligned}\].

8. This problem centers on finding antiderivatives for the basic trigonometric functions other than \(\sin(x)\) and \(\cos(x)\).

   1. Consider the indefinite integral \(\int \tan(x) \, dx\). By rewriting the integrand as \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) and identifying an appropriate function-derivative pair, make a \(u\)-substitution and hence evaluate \(\int \tan(x) \, dx\).

   2. In a similar way, evaluate \(\int \cot(x) \, dx\).

   3. Consider the indefinite integral \[\begin{aligned}\end{aligned}\]. Evaluate this integral using the substitution \(u = \sec(x) + \tan(x)\).

   4. Simplify the integrand in (c) by factoring the numerator. What is a far simpler way to write the integrand?

   5. Combine your work in (c) and (d) to determine \(\int \sec(x) \, dx\).

   6. Using (c)-(e) as a guide, evaluate \(\int \csc(x) \, dx\).

   Umsebenzi Icebo

   Bonisa impendulo

   1. We let \(u = \cos(x)\), so \(du = -\sin(x) \ dx\). This change of variable transforms the integral into one we can integrate as follows: \[\begin{aligned}\int \tan(x) \, dx &= \int \frac{\sin(x)}{\cos(x)} \ dx \\ &= -\int \frac{1}{u} \ du \\ &= -\ln(|u|) + C \\ &= -\ln(|\cos(x)| + C \\ &= \ln\left(|\cos(x)|^{-1}\right) + C \\ &= \ln\left(|\sec(x)|\right) + C\end{aligned}\]. Note that we used the property of the logarithm that \(p \ln (z) = \ln(z^p)\) as well as the fact that \(\frac{1}{\cos(x)} = \sec(x)\) in order to simplify the result.

   2. We let \(u = \sin(x)\), so \(du = \cos(x) \ dx\). This change of variable transforms the integral into one we can integrate as follows: \[\begin{aligned}\int \cot(x) \, dx &= \int \frac{\cos(x)}{\sin(x)} \ dx \\ &= \int \frac{1}{u} \ du \\ &= \ln(|u|) + C \\ &= \ln(|\sin(x)| + C \\ &= -\ln\left(|\sin(x)|^{-1}\right) + C \\ &= -\ln\left(|\csc(x)|\right) + C\end{aligned}\]. Note that here we used the fact that \(\ln(z) = (-1)(-1)\ln(z) = (-1) \ln(z^{-1})\) in order to simplify the result.

   3. We let \(u = \sec(x) + \tan(x)\), so it follows that \(du = \left(\sec(x)\tan(x)+\sec^2(x)\right) \ dx\). This change of variable shows that the basic structure of the integrand is simple and can be integrated as follows: \[\begin{aligned}\int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx &= \int \frac{1}{u} \ du \\ &= \ln(|u|) + C \\ &= \ln\left(|\sec(x) + \tan(x)|\right) + C\end{aligned}\].

   4. The numerator of \(\frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)}\) can be rewritten by removing the common factor of \(\sec(x)\) to obtain \[\begin{aligned}\end{aligned}\]. The integrand can then be simplified to \[\begin{aligned}\end{aligned}\].

   5. Using the results of (c) and (d), we see that \[\begin{aligned}\end{aligned}\].

   6. Following (c)-(e) we can evaluate \(int \csc(x) \, dx\) as follows. First, rewrite \(\csc(x)\) by multiplying by \(1 = \frac{\csc(x) + \cot(x)}{\csc(x) + \cot(x)}\), so that \[\begin{aligned}\end{aligned}\]. Now we evaluate the last integral in the preceding equation with the substitution \(u = \csc(x) + \cot(x)\). Then \(du = -\left(\csc(x)\cot(x)+\csc^2(x)\right) \ dx\). This change of variable transforms the integral so that we can integrate: \[\begin{aligned}\int \frac{\csc^2(x) + \csc(x) \cot(x)}{\csc(x) + \cot(x)} \, dx &= - \int \frac{1}{u} \ dx \\ &= -\ln(|u|) + C \\ &= -\ln\left(|\csc(x) + \cot(x)|\right) + C\end{aligned}\]. Connecting the equalities we have established in our work, we have shown that \[\begin{aligned}\end{aligned}\].

9. Consider the indefinite integral \(\int x \sqrt{x-1} \, dx\).

   1. At first glance, this integrand may not seem suited to substitution due to the presence of \(x\) in separate locations in the integrand. Nonetheless, using the composite function \(\sqrt{x-1}\) as a guide, let \(u = x-1\). Determine expressions for both \(x\) and \(dx\) in terms of \(u\).

   2. Convert the given integral in \(x\) to a new integral in \(u\).

   3. Evaluate the integral in (b) by noting that \(\sqrt{u} = u^{1/2}\) and observing that it is now possible to rewrite the integrand in \(u\) by expanding through multiplication.

   4. Evaluate each of the integrals \(\int x^2 \sqrt{x-1} \, dx\) and \(\int x \sqrt{x^2 - 1} \, dx\). Write a paragraph to discuss the similarities among the three indefinite integrals in this problem and the role of substitution and algebraic rearrangement in each.

   Umsebenzi Icebo

   Bonisa impendulo

   1. Letting \(u = x-1\), we find that \(x = u+1\) and thus \(dx = du\).

   2. Using the change of variables in (a), \[\begin{aligned}\end{aligned}\].

   3. Using the fact that \(\sqrt{u} = u^{1/2}\) and distributing, we can evaluate the integral as follows: \[\begin{aligned}\int x \sqrt{x-1} \, dx &= \int (u+1) u^{1/2} \, du \\ &= \int \left(u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\].

   4. First, for \(\int x^2 \sqrt{x-1} \, dx\), we let \(u = x-1\) so that \(x = u+1\) and \(dx = du\). From there, our work is very similar to what we did in (b): \[\begin{aligned}\int x^2 \sqrt{x-1} \, dx &= \int (u+1)^2 u^{1/2} \, du \\ &= \int \left(u^2 + 2u + 1 \right) u^{1/2} \, du \\ &= \int \left(u^{5/2} + 2u^{3/2} + u^{1/2} \right) \, du \\ &= \frac{2}{7} u^{7/2} + \frac{4}{5} u^{5/2} + \frac{2}{3} u^{3/2} + C \\ &= \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\end{aligned}\].

      For \(\int x \sqrt{x^2 - 1} \, dx\), despite the fact that the integral looks remarkably similar, the situation is quite different. We observe that with \(u = x^2 - 1\), it follows \(du = 2x \, dx\), so we have a standard function-derivative pair. Noting that \(x \, dx = \frac{1}{2} \, du\), it follows that \[\begin{aligned}\int x \sqrt{x^2 - 1} \, dx &= \int \sqrt{u} \cdot \frac{1}{2} \, du \\ &= \frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C \\ &= \frac{1}{3} u^{3/2} + C \\ &= \frac{1}{3} (x^2-1)^{3/2} + C\end{aligned}\].

      In the situation where we have \((x-a)\) under a square root and a polynomial outside the square root, the substitution \(u = x-a\) will allow us to convert the integrand into an expression we can integrate by changing the form of the radical from \(\sqrt{x-a}\) to \(\sqrt{u}\). From there we are able to distribute and integrate an expression that is essentially a sum of (fractional) powers of \(u\). In the situation where a polynomial of degree higher than \(1\) (such as \(x^2 - 1\)) is under the radical, we likely have to be fortunate enough to have the derivative of said polynomial (up to a constant) outside the radical in order to be able to integrate, as we were in the last example.

10. Consider the indefinite integral \(\int \sin^3(x) \, dx\).

    1. Explain why the substitution \(u = \sin(x)\) will not work to help evaluate the given integral.

    2. Recall the Fundamental Trigonometric Identity, which states that \(\sin^2(x) + \cos^2(x) = 1\). By observing that \(\sin^3(x) = \sin(x) \cdot \sin^2(x)\), use the Fundamental Trigonometric Identity to rewrite the integrand as the product of \(\sin(x)\) with another function.

    3. Explain why the substitution \(u = \cos(x)\) now provides a possible way to evaluate the integral in (b).

    4. Use your work in (a)-(c) to evaluate the indefinite integral \(\int \sin^3(x) \, dx\).

    5. Use a similar approach to evaluate \(\int \cos^3(x) \, dx\).

    Umsebenzi Icebo

    Bonisa impendulo

    1. If we let \(u = \sin(x)\), then \(du = \cos(x) \, dx\). Since there is no cosine function present in the integrand, we don't have the needed function-derivative pair present in order to transform the integral into one we can evaluate.

    2. Using the Fundamental Trigonometric Identity, we observe that \(\sin^2(x) = 1-\cos^2(x)\), and thus \[\begin{aligned}\end{aligned}\].

    3. By our work in (b), we see that \(\int \sin^3(x) \, dx = \int \sin(x) (1-\cos^2(x)) \, dx\), and in this integral we have the function-derivative pair that corresponds to \(u = \cos(x)\) and \(du = -\sin(x) \, dx\).

    4. Using this substitution and evaluating the resulting integral, \[\begin{aligned}\int \sin^3(x) \, dx &= \int \sin(x) (1-\cos^2(x)) \, dx \\ &= \int (-1)(1-u^2) \, du \\ &= \int (u^2-1) \, du \\ &= \frac{1}{3}u^3 - u + C \\ &= \frac{1}{3}\cos^3(x) - \cos(x) + C\end{aligned}\].

    5. Here we observe that \(\cos^3(x) = \cos(x) \cdot \cos^2(x) = \cos(x) (1-\sin^2(x))\), so letting \(u = \sin(x)\) and thus \(du = \cos(x) \, dx\), we have \[\begin{aligned}\int \cos^3(x) \, dx &= \int \cos(x) (1-\sin^2(x)) \, dx \\ &= \int (1-u^2) \, du \\ &= u - \frac{1}{3}u^3 + C \\ &= \sin(x) - \frac{1}{3}\sin^3(x) + C\end{aligned}\].

Ukugcina umsebenzi wakho

I akhawunti ekhululekileyo idibanisa amaphetshana kwi ncwadi nganye, irekhodi lento ogqibe ngayo, iingxaki zakho ezisoliweyo kwindawo enye, nomfundi onokuthi ubuze malunga nale phepha. IiMathematiki ngokwazo zivuliwe kubo bonke, bangeniswe okanye hayi.

Bhalisa Igama elithile

Iimpawu ezisetyenziswa apha

Nqakraza nasiphi na isibonakaliso sokuqonda okupheleleyo, umfanekiso, nokuba iileta zonke zithetha ntoni.

\int f(x)integral
Antiderivative (indefinite) or signed area from a to b (definite).\sqrt{x}square root, n-th root
The non-negative number whose square (n-th power) is x.\log_b xlogarithm, natural log
The exponent b must be raised to for x; ln uses base e.\sinsine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.f'(x)derivative
Instantaneous rate of change; slope of the graph.eEuler's number
2.71828…, the base whose exponential is its own derivative.\approxapproximately equal
Equal to the precision shown, not exactly.\leqless/greater than or equal
Inequalities that allow equality; < and > exclude it.\pipi
Ratio of a circle's circumference to its diameter, 3.14159…\inftyinfinity
Not a number: "grows without bound" in limits and intervals.\sum_{k=1}^{n} a_ksummation
Add a_k for k = 1 up to n.\lim_{x \to a} f(x)limit
The value f(x) approaches as x approaches a.y'first and second derivative of y
Prime notation for derivatives with respect to x (or t).Carbitrary constants
Constants of integration fixed by initial conditions.

Imibuzo abantu bebuza

What is a derivative in one sentence?

The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.

Iindawo zephepha zitshintshiwe Boelkins, Active Calculus (CC BY-SA 4.0), Wikipedia (CC BY-SA 4.0). Igqityiwe kwaye iphinde yacaciswa apha; iimpazamo zethu.

IiNkqubo Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests