8 Techniques of Integration

8.1 Integration by Parts

Here’s a simple integral that we can’t yet evaluate:


It’s a simple matter to take the derivative of the integrand using the Product Rule, but there is no Product Rule for integrals. However, this section introduces Integration by Parts, a method of integration that is based on the Product Rule for derivatives. It will enable us to evaluate this integral.

The Product Rule says that if u and v are functions of x, then (uv)=uv+uv. For simplicity, we’ve written u for u(x) and v for v(x). Suppose we integrate both sides with respect to x. This gives


By the Fundamental Theorem of Calculus, the left side integrates to uv. The right side can be broken up into two integrals, and we have


Solving for the second integral we have


Using differential notation, we can write

u=dudxv=dvdx    du=udxdv=vdx.

Thus, the equation above can be written as follows:


This is the Integration by Parts formula. For reference purposes, we state this in a theorem.

Theorem 8.1.1 Integration by Parts

Let u and v be differentiable functions of x on an interval I containing a and b. Then


and applying FTC part 2 we have


Let’s try an example to understand our new technique.

Example 8.1.1 Integrating using Integration by Parts

Evaluate xcosxdx.

SolutionThe key to Integration by Parts is to identify part of the integrand as “u” and part as “dv.” Regular practice will help one make good identifications, and later we will introduce some principles that help. For now, let u=x and dv=cosxdx.

It is generally useful to make a small table of these values.

u=xdv=cosxdxdu=?v=?    u=xdv=cosxdxdu=dxv=sinx

Right now we only know u and dv as shown on the left; on the right we fill in the rest of what we need. If u=x, then du=dx. Since dv=cosxdx, v is an antiderivative of cosx, so v=sinx.

Now substitute all of this into the Integration by Parts formula, giving


We can then integrate sinx to get -cosx+C and overall our answer is


We have two important notes here: (1) notice how the antiderivative contains the product, xsinx. This product is what makes integration by parts necessary. And (2) antidifferentiating dv does result in v+C. The intermediate +Cs are all added together and represented by one +C in the final answer.

The example above demonstrates how Integration by Parts works in general. We try to identify u and dv in the integral we are given, and the key is that we usually want to choose u and dv so that du is simpler than u and v is hopefully not too much more complicated than dv. This will mean that the integral on the right side of the Integration by Parts formula, vdu will be simpler to integrate than the original integral udv.

In the example above, we chose u=x and dv=cosxdx. Then du=dx was simpler than u and v=sinx is no more complicated than dv. Therefore, instead of integrating xcosxdx, we could integrate sinxdx, which we knew how to do.

If we had chosen u=cosx and dv=xdx, so that du=-sinxdx and v=12x2, then


We then need to integrate x2sinx, which is more complicated than our original integral, making this an unproductive choice.

We now consider another example.

Example 8.1.2 Integrating using Integration by Parts

Evaluate xexdx.

SolutionNotice that x becomes simpler when differentiated and ex is unchanged by differentiation or integration. This suggests that we should let u=x and dv=exdx:

u=xdv=exdxdu=?v=?    u=xdv=exdxdu=dxv=ex

The Integration by Parts formula gives


The integral on the right is simple; our final answer is


Note again how the antiderivatives contain a product term.

Example 8.1.3 Integrating using Integration by Parts

Evaluate x2cosxdx.

SolutionLet u=x2 instead of the trigonometric function, hence dv=cosxdx. Then du=2xdx and v=sinx as shown below.

u=x2dv=cosxdxdu=?v=?    u=x2dv=cosxdxdu=2xdxv=sinx

The Integration by Parts formula gives


At this point, the integral on the right is indeed simpler than the one we started with, but to evaluate it, we need to do Integration by Parts again. Here we choose u=2x and dv=sinxdx and fill in the rest below.

u=2xdv=sinxdxdu=?v=?    u=2xdv=sinxdxdu=2dxv=-cosx

This means that


The integral all the way on the right is now something we can evaluate. It evaluates to -2sinx. Then going through and simplifying, being careful to keep all the signs straight, our answer is

Example 8.1.4 Integrating using Integration by Parts

Evaluate excosxdx.

SolutionThis is a classic problem. In this particular example, one can let u be either cosx or ex; we choose u=ex and hence dv=cosxdx. Then du=exdx and v=sinx as shown below.

u=exdv=cosxdxdu=?v=?    u=exdv=cosxdxdu=exdxv=sinx

Notice that du is no simpler than u, going against our general rule (but bear with us). The Integration by Parts formula yields


The integral on the right is not much different from the one we started with, so it seems like we have gotten nowhere. Let’s keep working and apply Integration by Parts to the new integral. So what should we use for u and dv this time? We may feel like letting the trigonometric function be dv and the exponential be u was a bad choice last time since we still can’t integrate the new integral. However, if we let u=sinx and dv=exdx this time we will reverse what we just did, taking us back to the beginning. So, we let u=ex and dv=sinxdx. This leads us to the following:

u=exdv=sinxdxdu=?v=?    u=exdv=sinxdxdu=exdxv=-cosx

The Integration by Parts formula then gives:

excosxdx =exsinx-(-excosx--excosxdx)

It seems we are back right where we started, as the right hand side contains excosxdx. But this is actually a good thing.

Add excosxdx to both sides. This gives

2excosxdx =exsinx+excosx
Now divide both sides by 2:
excosxdx =12(exsinx+excosx).

Simplifying a little and adding the constant of integration, our answer is thus

Example 8.1.5 Using Integration by Parts: antiderivative of lnx

Evaluate lnxdx.

SolutionOne may have noticed that we have rules for integrating the familiar trigonometric functions and ex, but we have not yet given a rule for integrating lnx. That is because lnx can’t easily be integrated with any of the rules we have learned up to this point. But we can find its antiderivative by a clever application of Integration by Parts. Set u=lnx and dv=dx. This is a good strategy to learn as it can help in other situations. This determines du=(1/x)dx and v=x as shown below.

u=lnxdv=dxdu=?v=?    u=lnxdv=dxdu=1/xdxv=x

Putting this all together in the Integration by Parts formula, things work out very nicely:

lnxdx =xlnx-x1xdx
Example 8.1.6 Using Integration by Parts: antiderivative of tan-1x

Evaluate tan-1xdx.

SolutionThe same strategy of dv=dx that we used above works here. Let u=tan-1x and dv=dx. Then du=1/(1+x2)dx and v=x. The Integration by Parts formula gives


The integral on the right can be solved by substitution. Taking t=1+x2, we get dt=2xdx. The integral then becomes


The integral on the right evaluates to ln|t|+C, which becomes ln(1+x2)+C. Therefore, the answer is


Since 1+x2>0, we do not need to include the absolute value in the ln(1+x2) term.

Substitution Before Integration

When taking derivatives, it was common to employ multiple rules (such as using both the Quotient and the Chain Rules). It should then come as no surprise that some integrals are best evaluated by combining integration techniques. In particular, here we illustrate making an “unusual” substitution first before using Integration by Parts.

Example 8.1.7 Integration by Parts after substitution

Evaluate cos(lnx)dx.

SolutionThe integrand contains a composition of functions, leading us to think Substitution would be beneficial. Letting u=lnx, we have du=1/xdx. This seems problematic, as we do not have a 1/x in the integrand. But consider:


Since u=lnx, we can use inverse functions to solve for x=eu. Therefore we have that

dx =xdu

We can thus replace lnx with u and dx with eudu. Thus we rewrite our integral as


We evaluated this integral in Example 8.1.4. Using the result there, we have:

cos(lnx)dx =eucosudu

Definite Integrals and Integration By Parts

So far we have focused only on evaluating indefinite integrals. Of course, we can use Integration by Parts to evaluate definite integrals as well, as Theorem 8.1.1 states. We do so in the next example.

Example 8.1.8 Definite integration using Integration by Parts

Evaluate 12x2lnxdx.

SolutionTo simplify the integral we let u=lnx and dv=x2dx. We then get du=(1/x)dx and v=x3/3 as shown below.

u=lnxdv=x2dxdu=?v=?    u=lnxdv=x2dxdu=1/xdxv=x3/3

This may seem counterintuitive since the power on the algebraic factor has increased (v=x3/3), but as we see this is a wise choice:

12x2lnxdx =x33lnx|12-12x331xdx

In general, Integration by Parts is useful for integrating certain products of functions, like xexdx or x3sinxdx. It is also useful for integrals involving logarithms and inverse trigonometric functions.

As stated before, integration is generally more difficult than differentiation. We are developing tools for handling a large array of integrals, and experience will tell us when one tool is preferable/necessary over another. For instance, consider the three similar-looking integrals

xexdx,xex2dx  and  xex3dx.

While the first is calculated easily with Integration by Parts, the second is best approached with Substitution. Taking things one step further, the third integral has no answer in terms of elementary functions, so none of the methods we learn in calculus will get us the exact answer. We will learn how to approximate this integral in Chapter 9

Integration by Parts is a very useful method, second only to substitution. In the following sections of this chapter, we continue to learn other integration techniques. The next section focuses on handling integrals containing trigonometric functions.

Exercises 8.1


Terms and Concepts

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    T/F: Integration by Parts is useful in evaluating integrands that contain products of functions.

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    T/F: Integration by Parts can be thought of as the “opposite of the Chain Rule.”


In Exercises 3–36., evaluate the given indefinite integral.

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In Exercises 37–42., evaluate the indefinite integral after first making a substitution.

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In Exercises 43–52., evaluate the definite integral. Note: the corresponding indefinite integrals appear in Exercises 3.12..

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  1. 53.
    (a) For n2 show that 0π/2sinnxdx=n-1n0π/2sinn-2xdx. Hint: Begin by writing sinnx as (sinn-1x)sinx and using Integration by Parts. (b) For k1 show that 0π/2sin2kxdx =135(2k-1)246(2k)π2  and 0π/2sin2k+1xdx =246(2k)1357(2k+1).
  2. 54.
    Find the volume of the solid of revolution obtained by rotating the region bounded by y=0, y=lnx, x=1, and x=e: (a) About the x-axis, using the disk method. (b) About the y-axis, using the shell method.
  3. 55.
    Let f(x)=x for -πx<π and extend this function so that it is periodic with period 2π. This function is known as a sawtooth wave and looks like -7-6-5-4-3-2-11234567-5-4-3-2-112345xy For a positive integer n, define bn=1π-ππf(x)sin(nx)dx. (a) Find bn. (b) Graph n=1Nbnsin(nx) for various values of N. What do you observe?
  4. 56.
    Let f(x)={-x-π-πx<-π2x-π2x<π2π-xπ2x<π and extend this function so that it is periodic with period 2π. This function is known as a triangle wave and looks like -7-6-5-4-3-2-11234567-22xy For a positive integer n, define bn=1π-ππf(x)sin(nx)dx. (a) Find bn. (b) Graph n=1Nbnsin(nx) for various values of N. What do you observe?
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