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Integration by Parts

Recognize when to use integration by parts.

The Integration-by-Parts Formula

If, \(h(x)=f(x)g(x),\) then by using the product rule, we obtain \({h}^{'}(x)={f}^{'}(x)g(x)+{g}^{'}(x)f(x).\) Although at first it may seem counterproductive, let’s now integrate both sides of this equation: \(\int {h}^{'}(x)dx=\int (g(x){f}^{'}(x)+f(x){g}^{'}(x))dx.\)

This gives us

\[h(x)=f(x)g(x)=\int g(x){f}^{'}(x)dx+\int f(x){g}^{'}(x)dx.\]

Now we solve for \(\int f(x){g}^{'}(x)dx:\)

\[\int f(x){g}^{'}(x)dx=f(x)g(x)-\int g(x){f}^{'}(x)dx.\]

By making the substitutions \(u=f(x)\) and \(v=g(x),\) which in turn make \(du={f}^{'}(x)dx\) and \(dv={g}^{'}(x)dx,\) we have the more compact form

\[\int u\ dv=uv-\int v\ du.\]

The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.

Example

Try it.

Use integration by parts with \(u=x\) and \(dv=\text{sin}\ x\ dx\) to evaluate \(\int x\ \text{sin}\ x\ dx.\)

Solution

By choosing \(u=x,\) we have \(du=1dx.\) Since \(dv=\text{sin}\ x\ dx,\) we get \(v=\int \text{sin}\ x\ dx=\text{-}\text{cos}\ x.\) It is handy to keep track of these values as follows:

\[\begin{array}{lllllll}u & = & x & & dv & = & \text{sin}\ x\ dx \\ du & = & 1dx & & v & = & \int \text{sin}\ x\ dx=\text{-}\text{cos}\ x.\end{array}\]

Applying the integration-by-parts formula results in

\[\begin{array}{llll}\int x\ \text{sin}\ x\ dx & =(x)(\text{-}\text{cos}\ x)-\int (\text{-}\text{cos}\ x)(1dx) & & \text{Substitute}. \\ & =\text{-}x\ \text{cos}\ x+\int \text{cos}\ x\ dx & & \text{Simplify}. \\ & =\text{-}x\ \text{cos}\ x+\text{sin}\ x+C. & & \text{Use}\ \int \text{cos}\ x\ dx=\text{sin}\ x+C.\end{array}\]

The natural question to ask at this point is: How do we know how to choose \(u\) and \(dv?\) Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This mnemonic serves as an aid in determining an appropriate choice for \(u.\)

Condensed — the full section is in OpenStax Calculus Volume 2.

Integration by Parts for Definite Integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Example

Try it.

Find the area of the region bounded above by the graph of \(y={\text{tan}}^{-1}x\) and below by the \(x\)-axis over the interval \([0,1].\)

Solution

This region is shown in . To find the area, we must evaluate \(\int _{0}^{1}{\text{tan}}^{-1}x\ dx.\)

For this integral, let’s choose \(u={\text{tan}}^{-1}x\) and \(dv=dx,\) thereby making \(du=\frac{1}{{x}^{2}+1}dx\) and \(v=x.\) After applying the integration-by-parts formula () we obtain

\[\text{Area}=x\ {\text{tan}}^{-1}{x|}_{0}^{1}-\int _{0}^{1}\frac{x}{{x}^{2}+1}dx.\]

Use u-substitution to obtain

\[\int _{0}^{1}\frac{x}{{x}^{2}+1}dx=\frac{1}{2}\text{ln}|{x}^{2}+1{|}_{0}^{1}.\]

Thus,

\[\text{Area}=x\ {\tan }^{-1}x{|}_{0}^{1}-\frac{1}{2}\ln |{x}^{2}+1|{|}_{0}^{1}=\frac{\pi }{4}-\frac{1}{2}\ln \ 2.\]

At this point it might not be a bad idea to do a “reality check” on the reasonableness of our solution. Since \(\frac{\pi }{4}-\frac{1}{2}\text{ln}\ 2\approx 0.4388,\) and from we expect our area to be slightly less than 0.5, this solution appears to be reasonable.

Example

Try it.

Find the volume of the solid obtained by revolving the region bounded by the graph of \(f(x)={e}^{\text{-}x},\) the x-axis, the y-axis, and the line \(x=1\) about the y-axis.

Solution

The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).

To find the volume using shells, we must evaluate \(2\pi {\int }_{0}^{1}x{e}^{\text{-}x}dx.\) To do this, let \(u=x\) and \(dv={e}^{\text{-}x}.\) These choices lead to \(du=dx\) and \(v={\int }^{\text{}}{e}^{\text{-}x}=\text{-}{e}^{\text{-}x}.\) Substituting into , we obtain

\[\begin{array}{llll}\text{Volume} & =2\pi \int _{0}^{1}x{e}^{\text{-}x}dx=2\pi (\text{-}x{e}^{\text{-}x}{|}_{0}^{1}+\int _{0}^{1}{e}^{\text{-}x}dx) & & \text{Use integration by parts}. \\ & =-2\pi x{e}^{\text{-}x}{|}_{0}^{1}-2\pi {e}^{\text{-}x}{|}_{0}^{1} & & \text{Evaluate}\ \int _{0}^{1}{e}^{\text{-}x}dx=\text{-}{e}^{\text{-}x}{|}_{0}^{1}. \\ & =2\pi -\frac{4\pi }{e}. & & \text{Evaluate and simplify}.\end{array}\]

Key Concepts

The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
Integration by parts applies to both definite and indefinite integrals.

Key Equations

Integration by parts formula	\(\int u\ dv=uv-\int v\ du\)
Integration by parts for definite integrals	\({\int }_{a}^{b}u\ dv={uv\|}_{a}^{b}-{\int }_{a}^{b}v\ du\)

Integration by Parts

In using the technique of integration by parts, you must carefully choose which expression is u. For each of the following problems, use the guidelines in this section to choose u. Do not evaluate the integrals.

Find the integral by using the simplest method. Not all problems require integration by parts.

Compute the definite integrals. Use a graphing utility to confirm your answers.

Derive the following formulas using the technique of integration by parts. Assume that n is a positive integer. These formulas are called reduction formulas because the exponent in the x term has been reduced by one in each case. The second integral is simpler than the original integral.

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique used to perform the integration without actually doing the problem.

Sketch the region bounded above by the curve, the x-axis, and \(x=1,\) and find the area of the region. Provide the exact form or round answers to the number of places indicated.

Find the volume generated by rotating the region bounded by the given curves about the specified line. Express the answers in exact form or approximate to the number of decimal places indicated.

Introduction

In Section, we learned the technique of \(u\)-substitution for evaluating indefinite integrals. For example, the indefinite integral \(\int x^3 \sin(x^4) \, dx\) is perfectly suited to \(u\)-substitution, because one factor is a composite function and the other factor is the derivative (up to a constant) of the inner function. Recognizing the algebraic structure of a function can help us to find its antiderivative.

Next we consider integrands with a different elementary algebraic structure: a product of basic functions. For instance, suppose we are interested in evaluating the indefinite integral \[\begin{aligned}\end{aligned}\].

The integrand is the product of the basic functions \(f(x) = x\) and \(g(x) = \sin(x)\). We know that it is relatively complicated to compute the derivative of the product of two functions, so we should expect that antidifferentiating a product should be similarly involved. Intuitively, we expect that evaluating \(\int x \sin(x) \, dx\) will involve somehow reversing the Product Rule.

To that end, in Preview Activity we refresh our understanding of the Product Rule and then investigate some indefinite integrals that involve products of basic functions.

Exploration

Reversing the Product Rule: Integration by Parts

Problem (c) in Preview Activity provides a clue to the general technique known as Integration by Parts, which comes from reversing the Product Rule. Recall that the Product Rule states that \[\begin{aligned}\end{aligned}\].

Integrating both sides of this equation indefinitely with respect to \(x\), we find \[\begin{aligned}\end{aligned}\].

On the left side of Equation, we have the indefinite integral of the derivative of a function. Temporarily omitting the constant that may arise, we have \[\begin{aligned}\end{aligned}\].

We solve for the first indefinite integral on the left to generate the rule \[\begin{aligned}\end{aligned}\].

Often we express Equation in terms of the variables \(u\) and \(v\), where \(u = f(x)\) and \(v = g(x)\). In differential notation, \(du = f'(x) \, dx\) and \(dv = g'(x) \, dx\), so we can state the rule for Integration by Parts in its most common form as follows:

\[\begin{aligned}\end{aligned}\].

To apply integration by parts, we look for a product of basic functions that we can identify as \(u\) and \(dv\). If we can antidifferentiate \(dv\) to find \(v\), and evaluating \(\int v \, du\) is not more difficult than evaluating \(\int u \, dv\), then this substitution usually proves to be fruitful. To demonstrate, we consider the following example.

The general technique of integration by parts involves trading the problem of integrating the product of two functions for the problem of integrating the product of two related functions. That is, we convert the problem of evaluating \(\int u \, dv\) to that of evaluating \(\int v \, du\). This clearly shapes our choice of \(u\) and \(v\). In Example, the original integral to evaluate was \(\int x \cos(x) \,dx\), and through the substitution provided by integration by parts, we were instead able to evaluate \(\int \sin(x) \cdot 1 \, dx\). Note that the original function \(x\) was replaced by its derivative, while \(\cos(x)\) was replaced by its antiderivative.

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

Some Subtleties with Integration by Parts

Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating \[\begin{aligned}\end{aligned}\].

Initially, this problem seems ill-suited to integration by parts, since there does not appear to be a product of functions present. But if we note that \(\arctan(x) = \arctan(x) \cdot 1\), and realize that we know the derivative of \(\arctan(x)\) as well as the antiderivative of \(1\), we see the possibility for the substitution \(u = \arctan(x)\) and \(dv = 1 \, dx\). We explore this substitution further in Activity.

In a related problem, consider \(\int t^3 \sin(t^2) \, dt\). Observe that there is a composite function present in \(\sin(t^2)\), but there is not an obvious function-derivative pair, as we have \(t^3\) (rather than simply \(t\)) multiplying \(\sin(t^2)\). In this problem we use both \(u\)-substitution and integration by parts. First we write \(t^3 = t \cdot t^2\) and consider the indefinite integral \[\begin{aligned}\end{aligned}\]. We let \(z = t^2\) so that \(dz = 2t \, dt\), and thus \(t \, dt = \frac{1}{2} \, dz\). (We are using the variable \(z\) to perform a \(z\)-substitution first so that we may then apply integration by parts.) Under this \(z\)-substitution, we now have \[\begin{aligned}\end{aligned}\].

The resulting integral can be evaluated by parts. This, too, is explored further in Activity.

These problems show that we sometimes must think creatively in choosing the variables for substitution in integration by parts, and that we may need to use substitution for an additional change of variables.

Using Integration by Parts Multiple Times

Integration by parts is well suited to integrating the product of basic functions, allowing us to trade a given integrand for a new one where one function in the product is replaced by its derivative, and the other is replaced by its antiderivative. The goal in this trade of \(\int u \, dv\) for \(\int v \, du\) is that the new integral be simpler to evaluate than the original one. Sometimes it is necessary to apply integration by parts more than once in order to evaluate a given integral.

Example

Evaluate \(\int t^2 e^t \, dt\).

Solution

Let \(u = t^2\) and \(dv = e^t \, dt\). Then \(du = 2t \, dt\) and \(v = e^t\), and thus \[\begin{aligned}\end{aligned}\].

The integral on the right side is simpler to evaluate than the one on the left, but it still requires integration by parts. Now letting \(u = 2t\) and \(dv = e^t \, dt\), we have \(du = 2\, dt\) and \(v = e^t\), so that \[\begin{aligned}\end{aligned}\].

(Note the parentheses, which remind us to distribute the minus sign to the entire value of the integral \(\int 2t e^t \, dt\).) The final integral on the right is a basic one; evaluating that integral and distributing the minus sign, we find \[\begin{aligned}\end{aligned}\].

Of course, even more than two applications of integration by parts may be necessary. In the preceding example, if the integrand had been \(t^3e^t\), we would have had to use integration by parts three times.

Next, we consider the slightly different scenario.

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

Evaluating Definite Integrals Using Integration by Parts

We can use the technique of integration by parts to evaluate a definite integral as well.

Example

Evaluate \[\begin{aligned}\end{aligned}\].

Solution

One option is to find an antiderivative (using indefinite integral notation) and then apply the Fundamental Theorem of Calculus to find that \[\begin{aligned}\int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp \left( -t \cos(t) + \sin(t) \right) \bigg\vert_0^{\pi/2} \\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right) \\ =\mathstrut \amp 1\end{aligned}\].

Alternatively, we can apply integration by parts and work with definite integrals throughout. With this method, we must remember to evaluate the product \(uv\) over the given limits of integration. Using the substitution \(u = t\) and \(dv = \sin(t) \, dt\), so that \(du = dt\) and \(v = -\cos(t)\), we write \[\begin{aligned}\int_0^{\pi/2} t\sin(t) \, dt =\mathstrut \amp -t \cos(t) \bigg\vert_0^{\pi/2} - \int_0^{\pi/2} (-\cos(t)) \, dt \\ =\mathstrut \amp -t \cos(t) \bigg\vert_0^{\pi/2} + \sin(t) \bigg\vert_0^{\pi/2} \\ =\mathstrut \amp \left( -\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2}) \right) - \left( -0 \cos(0) + \sin(0) \right) \\ =\mathstrut \amp 1\end{aligned}\].

As with any substitution technique, it is important to use notation carefully and completely, and to ensure that the end result makes sense.

When u-substitution and Integration by Parts Fail to Help

Both integration techniques we have discussed apply in relatively limited circumstances. It is not hard to find examples of functions for which neither technique produces an antiderivative; indeed, there are many, many functions that appear elementary but that do not have an elementary algebraic antiderivative. For instance, neither \(u\)-substitution nor integration by parts proves fruitful for the indefinite integrals \[\begin{aligned}\end{aligned}\]. While there are other integration techniques, some of which we will consider briefly, none of them enables us to find an algebraic antiderivative for \(e^{x^2}\) or \(x \tan(x)\). We do know from the Second Fundamental Theorem of Calculus that we can construct an integral antiderivative for each function; \(F(x) = \int_0^x e^{t^2} \, dt\) is an antiderivative of \(f(x) = e^{x^2}\), and \(G(x) = \int_0^{x} t \tan(t) \, dt\) is an antiderivative of \(g(x) = x \tan(x)\). But finding an elementary algebraic formula that doesn't involve integrals for either \(F\) or \(G\) turns out not only to be impossible through \(u\)-substitution or integration by parts, but indeed impossible altogether. Antidifferentiation is much harder in general than differentiation.

Summary

Through the method of integration by parts, we can evaluate indefinite integrals that involve products of basic functions such as \(\int x \sin(x) \, dx\) and \(\int x \ln(x) \, dx\). Using a substitution enables us to trade one of the functions in the product for its derivative, and the other for its antiderivative, in an effort to find a different product of functions that is easier to integrate.
If the algebraic structure of an integrand is a product of basic functions in the form \(\int f(x) g'(x) \, dx\), we can use the substitution \(u = f(x)\) and \(dv = g'(x) \,dx\) and apply the rule \[\begin{aligned}\end{aligned}\] to evaluate the original integral \(\int f(x) g'(x) \, dx\) by instead evaluating \[\begin{aligned}\end{aligned}\].
When deciding to integrate by parts, we have to select both \(u\) and \(dv\). That selection is guided by the overall principle that the new integral \(\int v \, du\) not be more difficult than the original integral \(\int u \, dv\). In addition, it is often helpful to recognize if one of the functions present is much easier to differentiate than antidifferentiate (such as \(\ln(x)\)), in which case that function often is best assigned the variable \(u\). In addition, \(dv\) must be a function that we can antidifferentiate.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Use integration by parts with \(u=x\) and \(dv=\text{sin}\ x\ dx\) to evaluate \(\int x\ \text{sin}\ x\ dx.\)

Paljasta vastaus

By choosing \(u=x,\) we have \(du=1dx.\) Since \(dv=\text{sin}\ x\ dx,\) we get \(v=\int \text{sin}\ x\ dx=\text{-}\text{cos}\ x.\) It is handy to keep track of these values as follows:
\[\begin{array}{lllllll}u & = & x & & dv & = & \text{sin}\ x\ dx \\ du & = & 1dx & & v & = & \int \text{sin}\ x\ dx=\text{-}\text{cos}\ x.\end{array}\]
Applying the integration-by-parts formula results in
\[\begin{array}{llll}\int x\ \text{sin}\ x\ dx & =(x)(\text{-}\text{cos}\ x)-\int (\text{-}\text{cos}\ x)(1dx) & & \text{Substitute}. \\ & =\text{-}x\ \text{cos}\ x+\int \text{cos}\ x\ dx & & \text{Simplify}. \\ & =\text{-}x\ \text{cos}\ x+\text{sin}\ x+C. & & \text{Use}\ \int \text{cos}\ x\ dx=\text{sin}\ x+C.\end{array}\]
Evaluate \({\int }^{\text{}}x{e}^{2x}dx\) using the integration-by-parts formula with \(u=x\) and \(dv={e}^{2x}dx.\)

Paljasta vastaus

\({\int }^{\text{}}x{e}^{2x}dx=\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C\)
Evaluate \(\int \frac{\text{ln}\ x}{{x}^{3}}dx.\)

Paljasta vastaus

Begin by rewriting the integral:
\[\int \frac{\text{ln}\ x}{{x}^{3}}dx={\int }^{\text{}}{x}^{-3}\text{ln}\ x\ dx.\]
Since this integral contains the algebraic function \({x}^{-3}\) and the logarithmic function \(\text{ln}\ x,\) choose \(u=\text{ln}\ x,\) since L comes before A in LIATE. After we have chosen \(u=\text{ln}\ x,\) we must choose \(dv={x}^{-3}dx.\)
Next, since \(u=\text{ln}\ x,\) we have \(du=\frac{1}{x}dx.\) Also, \(v={\int }^{\text{}}{x}^{-3}dx=-\frac{1}{2}{x}^{-2}.\) Summarizing,
\[\begin{array}{lllllll}u & = & \text{ln}\ x & & dv & = & {x}^{-3}dx \\ du & = & \frac{1}{x}dx & & v & = & {\int }^{\text{}}{x}^{-3}dx=-\frac{1}{2}{x}^{-2}.\end{array}\]
Substituting into the integration-by-parts formula () gives
\[\begin{array}{lllll}\int \frac{\text{ln}\ x}{{x}^{3}}dx & ={\int }^{\text{}}{x}^{-3}\text{ln}\ x\ dx=(\text{ln}\ x)(\text{-}\ \frac{1}{2}{x}^{-2})-{\int }^{\text{}}(\text{-}\ \frac{1}{2}{x}^{-2})(\frac{1}{x}dx) & & & \\ & =-\frac{1}{2}{x}^{-2}\text{ln}\ x+{\int }^{\text{}}\frac{1}{2}{x}^{-3}dx & & & \text{Simplify}. \\ & =-\frac{1}{2}{x}^{-2}\text{ln}\ x-\frac{1}{4}{x}^{-2}+C & & & \text{Integrate}. \\ & =-\frac{1}{2{x}^{2}}\text{ln}\ x-\frac{1}{4{x}^{2}}+C. & & & \text{Rewrite with positive integers.}\end{array}\]
Evaluate \({\int }^{\text{}}x\ \text{ln}\ x\ dx.\)

Paljasta vastaus

\(\frac{1}{2}{x}^{2}\text{ln}\ x-\frac{1}{4}{x}^{2}+C\)
Evaluate \({\int }^{\text{}}{x}^{2}{e}^{3x}dx.\)

Paljasta vastaus

Using LIATE, choose \(u={x}^{2}\) and \(dv={e}^{3x}dx.\) Thus, \(du=2x\ dx\) and \(v=\int {e}^{3x}dx=(\frac{1}{3}){e}^{3x}.\) Therefore,
\[\begin{array}{lllllll}u & = & {x}^{2} & & dv & = & {e}^{3x}dx \\ du & = & 2x\ dx & & v & = & \int {e}^{3x}dx=\frac{1}{3}{e}^{3x}.\end{array}\]
Substituting into produces
\[\int {x}^{2}{e}^{3x}dx=\frac{1}{3}{x}^{2}{e}^{3x}-\int \frac{2}{3}x{e}^{3x}dx.\]
We still cannot integrate \(\int \frac{2}{3}x{e}^{3x}dx\) directly, but the integral now has a lower power on \(x.\) We can evaluate this new integral by using integration by parts again. To do this, choose \(u=x\) and \(dv=\frac{2}{3}{e}^{3x}dx.\) Thus, \(du=dx\) and \(v=\int (\frac{2}{3}){e}^{3x}dx=(\frac{2}{9}){e}^{3x}.\) Now we have
\[\begin{array}{lllllll}u & = & x & & dv & = & \frac{2}{3}{e}^{3x}dx \\ du & = & dx & & v & = & \int \frac{2}{3}{e}^{3x}dx=\frac{2}{9}{e}^{3x}.\end{array}\]
Substituting back into the previous equation yields
\[{\int }^{\text{}}{x}^{2}{e}^{3x}dx=\frac{1}{3}{x}^{2}{e}^{3x}-(\frac{2}{9}x{e}^{3x}-{\int }^{\text{}}\frac{2}{9}{e}^{3x}dx).\]
After evaluating the last integral and simplifying, we obtain
\[\int {x}^{2}{e}^{3x}dx=\frac{1}{3}{x}^{2}{e}^{3x}-\frac{2}{9}x{e}^{3x}+\frac{2}{27}{e}^{3x}+C.\]
Evaluate \({\int }^{\text{}}{t}^{3}{e}^{{t}^{2}}dt.\)

Paljasta vastaus

If we use a strict interpretation of the mnemonic LIATE to make our choice of \(u,\) we end up with \(u={t}^{3}\) and \(dv={e}^{{t}^{2}}dt.\) Unfortunately, this choice won’t work because we are unable to evaluate \({\int }^{\text{}}{e}^{{t}^{2}}dt.\) However, since we can evaluate \({\int }^{\text{}}t{e}^{{t}^{2}}dt,\) we can try choosing \(u={t}^{2}\) and \(dv=t{e}^{{t}^{2}}dt.\) With these choices we have
\[\begin{array}{lllllll}u & = & {t}^{2} & & dv & = & t{e}^{{t}^{2}}dt \\ du & = & 2t\ dt & & v & = & {\int }^{\text{}}t{e}^{{t}^{2}}dt=\frac{1}{2}{e}^{{t}^{2}}.\end{array}\]
Thus, we obtain
\[\begin{array}{ll}{\int }^{\ }{t}^{3}{e}^{{t}^{2}}dt & =\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\int \frac{1}{2}{e}^{{t}^{2}}2tdt \\ & =\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\frac{1}{2}{e}^{{t}^{2}}+C.\end{array}\]
Evaluate \({\int }^{\text{}}\text{sin}(\text{ln}\ x)dx.\)

Paljasta vastaus

This integral appears to have only one function—namely, \(\text{sin}(\text{ln}\ x)\) —however, we can always use the constant function 1 as the other function. In this example, let’s choose \(u=\text{sin}(\text{ln}\ x)\) and \(dv=1dx.\) (The decision to use \(u=\text{sin}(\text{ln}\ x)\) is easy. We can’t choose \(dv=\ \text{sin}(\text{ln}\ x)dx\) because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, \(du=(1\text{/}x)\text{cos}(\text{ln}\ x)dx\) and \(v={\int }^{\text{}}1dx=x.\) After applying integration by parts to the integral and simplifying, we have
\[{\int }^{\text{}}\text{sin}(\text{ln}\ x)dx=x\ \text{sin}(\text{ln}\ x)-{\int }^{\text{}}\text{cos}(\text{ln}\ x)dx.\]
Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s see what happens when we apply integration by parts again. This time let’s choose \(u=\text{cos}(\text{ln}\ x)\) and \(dv=1dx,\) making \(du=\text{-}(1\text{/}x)\text{sin}(\text{ln}\ x)dx\) and \(v={\int }^{\text{}}1dx=x.\) Substituting, we have
\[{\int }^{\text{}}\text{sin}(\text{ln}\ x)dx=x\ \text{sin}(\text{ln}\ x)-(x\ \text{cos}(\text{ln}\ x)—{\int }^{\text{}}-\text{sin}(\text{ln}\ x)dx).\]
After simplifying, we obtain
\[{\int }^{\text{}}\text{sin}(\text{ln}\ x)dx=x\ \text{sin}(\text{ln}\ x)-x\ \text{cos}(\text{ln}\ x)-{\int }^{\text{}}\text{sin}(\text{ln}\ x)dx.\]
The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute \(I={\int }^{\text{}}\text{sin}(\text{ln}\ x)dx.\) Thus, the equation becomes
\[I=x\ \text{sin}(\text{ln}\ x)-x\ \text{cos}(\text{ln}\ x)-I.\]
First, add \(I\) to both sides of the equation to obtain
\[2I=x\ \text{sin}(\text{ln}\ x)-x\ \text{cos}(\text{ln}\ x).\]
Next, divide by 2:
\[I=\frac{1}{2}x\ \text{sin}(\text{ln}\ x)-\frac{1}{2}x\ \text{cos}(\text{ln}\ x).\]
Substituting \(I={\int }^{\text{}}\text{sin}(\text{ln}\ x)dx\) again, we have
\[{\int }^{\text{}}\text{sin}(\text{ln}\ x)dx=\frac{1}{2}x\ \text{sin}(\text{ln}\ x)-\frac{1}{2}x\ \text{cos}(\text{ln}\ x).\]
From this we see that \((1\text{/}2)x\ \text{sin}(\text{ln}\ x)-(1\text{/}2)x\ \text{cos}(\text{ln}\ x)\) is an antiderivative of \(\text{sin}(\text{ln}\ x)dx.\) For the most general antiderivative, add \(+C\text{:}\)
\[{\int }^{\text{}}\text{sin}(\text{ln}\ x)dx=\frac{1}{2}x\ \text{sin}(\text{ln}\ x)-\frac{1}{2}x\ \text{cos}(\text{ln}\ x)+C.\]
Evaluate \({\int }^{\text{}}{x}^{2}\text{sin}\ x\ dx.\)

Paljasta vastaus

\(\text{-}{x}^{2}\text{cos}\ x+2x\ \text{sin}\ x+2\ \text{cos}\ x+C\)
Find the area of the region bounded above by the graph of \(y={\text{tan}}^{-1}x\) and below by the \(x\)-axis over the interval \([0,1].\)

Paljasta vastaus

This region is shown in . To find the area, we must evaluate \(\int _{0}^{1}{\text{tan}}^{-1}x\ dx.\)
For this integral, let’s choose \(u={\text{tan}}^{-1}x\) and \(dv=dx,\) thereby making \(du=\frac{1}{{x}^{2}+1}dx\) and \(v=x.\) After applying the integration-by-parts formula () we obtain
\[\text{Area}=x\ {\text{tan}}^{-1}{x|}_{0}^{1}-\int _{0}^{1}\frac{x}{{x}^{2}+1}dx.\]
Use u-substitution to obtain
\[\int _{0}^{1}\frac{x}{{x}^{2}+1}dx=\frac{1}{2}\text{ln}|{x}^{2}+1{|}_{0}^{1}.\]
Thus,
\[\text{Area}=x\ {\tan }^{-1}x{|}_{0}^{1}-\frac{1}{2}\ln |{x}^{2}+1|{|}_{0}^{1}=\frac{\pi }{4}-\frac{1}{2}\ln \ 2.\]
At this point it might not be a bad idea to do a “reality check” on the reasonableness of our solution. Since \(\frac{\pi }{4}-\frac{1}{2}\text{ln}\ 2\approx 0.4388,\) and from we expect our area to be slightly less than 0.5, this solution appears to be reasonable.
Find the volume of the solid obtained by revolving the region bounded by the graph of \(f(x)={e}^{\text{-}x},\) the x-axis, the y-axis, and the line \(x=1\) about the y-axis.

Paljasta vastaus

The best option to solving this problem is to use the shell method. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).
To find the volume using shells, we must evaluate \(2\pi {\int }_{0}^{1}x{e}^{\text{-}x}dx.\) To do this, let \(u=x\) and \(dv={e}^{\text{-}x}.\) These choices lead to \(du=dx\) and \(v={\int }^{\text{}}{e}^{\text{-}x}=\text{-}{e}^{\text{-}x}.\) Substituting into , we obtain
\[\begin{array}{llll}\text{Volume} & =2\pi \int _{0}^{1}x{e}^{\text{-}x}dx=2\pi (\text{-}x{e}^{\text{-}x}{|}_{0}^{1}+\int _{0}^{1}{e}^{\text{-}x}dx) & & \text{Use integration by parts}. \\ & =-2\pi x{e}^{\text{-}x}{|}_{0}^{1}-2\pi {e}^{\text{-}x}{|}_{0}^{1} & & \text{Evaluate}\ \int _{0}^{1}{e}^{\text{-}x}dx=\text{-}{e}^{\text{-}x}{|}_{0}^{1}. \\ & =2\pi -\frac{4\pi }{e}. & & \text{Evaluate and simplify}.\end{array}\]
Evaluate \({\int }_{0}^{\pi \text{/}2}x\ \text{cos}\ x\ dx.\)

Paljasta vastaus

\(\frac{\pi }{2}-1\)
\(\int {x}^{3}{e}^{2x}dx\)

Paljasta vastaus

\(u={x}^{3}\)
\(\int {x}^{3}\text{ln}(x)dx\)
\(\int {y}^{3}\text{cos}\ ydy\)

Paljasta vastaus

\(u={y}^{3}\)
\(\int {x}^{2}\text{arctan}\ x\ dx\)
\(\int {e}^{3x}\text{sin}(2x)dx\)

Paljasta vastaus

\(u=\text{sin}(2x)\)
\(\int v\ \text{sin}\ vdv\)
\(\int \text{ln}\ x\ dx\) (Hint: \(\int \text{ln}\ x\ dx\) is equivalent to \(\int 1\cdot \text{ln}(x)dx.)\)

Paljasta vastaus

\(\text{-}x+x\ \text{ln}\ x+C\)
\(\int x\ \text{cos}\ x\ dx\)
\(\int {\text{tan}}^{-1}x\ dx\)

Paljasta vastaus

\(x\ {\text{tan}}^{-1}x-\frac{1}{2}\text{ln}(1+{x}^{2})+C\)
\(\int {x}^{2}{e}^{x}dx\)
\(\int x\ \text{sin}(2x)dx\)

Paljasta vastaus

\(-\frac{1}{2}x\ \text{cos}(2x)+\frac{1}{4}\text{sin}(2x)+C\)
\(\int x{e}^{4x}dx\)
\(\int x{e}^{\text{-}x}dx\)

Paljasta vastaus

\({e}^{\text{-}x}(-1-x)+C\)
\(\int x\ \text{cos}\ 3x\ dx\)
\(\int {x}^{2}\text{cos}\ x\ dx\)

Paljasta vastaus

\(2x\ \text{cos}\ x+(-2+{x}^{2})\text{sin}\ x+C\)
\(\int x\ \text{ln}\ x\ dx\)
\(\int \text{ln}(2x+1)dx\)

Paljasta vastaus

\(\frac{1}{2}(1+2x)(-1+\text{ln}(1+2x))+C\)
\(\int {x}^{2}{e}^{4x}dx\)
\(\int {e}^{x}\text{sin}\ x\ dx\)

Paljasta vastaus

\(\frac{1}{2}{e}^{x}(\text{-}\text{cos}\ x+\text{sin}\ x)+C\)
\(\int {e}^{x}\text{cos}\ x\ dx\)
\(\int x{e}^{\text{-}{x}^{2}}dx\)

Paljasta vastaus

\(-\frac{{e}^{\text{-}{x}^{2}}}{2}+C\)
\(\int {x}^{2}{e}^{\text{-}x}dx\)
\(\int \text{sin}(\text{ln}(2x))dx\)

Paljasta vastaus

\(-\frac{1}{2}x\ \text{cos}[\text{ln}(2x)]+\frac{1}{2}x\ \text{sin}[\text{ln}(2x)]+C\)
\(\int cos(\text{ln}\ x)dx\)
\(\int {(\text{ln}\ x)}^{2}dx\)

Paljasta vastaus

\(2x-2x\ \text{ln}\ x+x{(\text{ln}\ x)}^{2}+C\)
\(\int \text{ln}({x}^{2})dx\)
\(\int {x}^{2}\text{ln}\ x\ dx\)

Paljasta vastaus

\((\text{-}\ \frac{{x}^{3}}{9}+\frac{1}{3}{x}^{3}\text{ln}\ x)+C\)
\(\int {\text{sin}}^{-1}x\ dx\)
\(\int {\text{cos}}^{-1}(2x)dx\)

Paljasta vastaus

\(-\frac{1}{2}\sqrt{1-4{x}^{2}}+x\ {\text{cos}}^{-1}(2x)+C\)

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\approx

approximately equal
Equal to the precision shown, not exactly.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sum_{k=1}^{n} a_k

summation
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.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Integration by Parts

Recognize when to use integration by parts.
Use the integration-by-parts formula to solve integration problems.
Use the integration-by-parts formula for definite integrals.
The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
Integration by parts applies to both definite and indefinite integrals.
Using parts, letting
Substitution, letting

Questions people ask

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.

Kokeile omaasi

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Lisää Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests