Let and .

Then and .

Thus

We will compute integrals using the integration by parts technique.

### 1 INTRODUCTION

In this section, we are interested in integrating products. We have some experience
with this from our exposure to -substitution problems such as In this scenario, the
factor of in the integrand is of great significance as it is the derivative of which is the
inside of the composition . We will now turn to problems where the factors in
the product making up the integrand are unrelated in this or any other
significant way, i.e., they are functionally independent of one another. We will
effectively create a product rule for integrals and interestingly, we will use the
product rule for derivatives to help us. One stark difference between this
product rule for integrals and the product rule for derivatives, known as
**integration by parts** (IBP), is that IBP does not give us a final answer.
Instead, integration by parts simply transforms our problem into another,
hopefully easier one, which we then have to solve. In this last respect, IBP
is similar to -substitution. The following integrals can be computed using
IBP:

### 2 IBP Formula

Recall the product rule: (where and are functions of ). Integrating both sides of this equation with respect to gives where the has intentionally been suppressed from the notation for simplicity. Now the integral on the left hand side is simply by the anti-differentiation definition of the integral symbol. So, Subtracting the second integral from both sides gives which can be condensed as the famous integration by parts formula.

- 1.
- To use the IBP formula, the original problem must be written in the form . This means that and must be declared (similar to declaring in -substitution), and then we compute by differentiating , and by anti-differentiating .
- 2.
- The constant of integration does not appear on the right hand side of the IBP formula (it mysteriously disappeared in the above derivation) because it was taken into account by the constant of integration implied in the integral on the right hand side, .
- 3.
- The IBP formula does not give a final answer to a problem. It replaces one integral with another another integral (which still has to be computed).

We will now use the integration by parts (IBP) formula to compute integrals involving products of unrelated factors, like the ones shown above.

The integrand is the product of two factors, and . We consider one factor to differentiate and the other factor to anti-differentiate. The factor we prefer to differentiate is since the derivative of is a constant and constants are easy to deal with in integrals. As for the factor, whether we differentiate it or anti-differentiate it makes little difference. In fact, only a change in sign occurs, as Thus to apply the IBP formula, we let Next, we compute by differentiating and we compute by anti-differentiating :

We have found an answer, but at this stage we should prove that it is the correct answer by differentiating it. The result should be . Note that the differentiation requires the product rule on the term: . We have which proves that our answer was correct.

Let and .

Then and . Next, the IBP formula gives \begin{align*} \int 3x\sin (4x) \;dx &= \overbrace{3x}^{u}\overbrace{\left [-\frac 14 \cos (4x)\right ]}^{v} - \int \overbrace{-\frac 14 \cos (4x)}^{v} \cdot \overbrace{3 \; dx}^{du} \\ &= -\frac 34 x\cos (4x) + \frac 34 \int \cos (4x) \; dx \\ &= -\frac 34 x\cos (4x) + \frac 34 \cdot \frac 14 \sin (4x) + C\\ &= -\frac 34 x\cos (4x) + \frac{3}{16}\sin (4x) + C. \end{align*}

Next, the IBP formula gives \begin{align*} \int x^2\sin (3x) \;dx &= x^2\left [-\frac 13 \cos (3x)\right ] - \int -\frac 13 \cos (3x) \cdot 2x \; dx \\ &= -\frac 13 x^2\cos (3x) + \frac 13 \int 2x\cos (3x) \; dx\\ \text{we now use IBP} & \text{ again with } \;\; u = 2x \;\text{ and } \; dv = \cos (3x) \, dx\\ & \qquad \qquad \quad du = 2 \, dx \; \text{ and } \; v=\frac 13 \sin (3x)\\ &= -\frac 13 x^2\cos (3x) + \frac 13\left [\frac 23x\sin (3x) - \int \frac 13 \sin (3x)\cdot 2 \; dx\right ]\\ &= -\frac 13 x^2\cos (3x) + \frac 29x\sin (3x) - \frac 29 \int \sin (3x) \; dx\\ &= -\frac 13 x^2\cos (3x) + \frac 29x\sin (3x) + \frac{2}{27} \cos (3x) + C. \end{align*}

### 3 Tabular Integration

Integrals of the form require IBP times when is a sine, cosine or exponential
function. A notational shortcut called **tabular integration** can be used to
streamline the process of using IBP multiple times. As an example, let’s revisit the
previous example:

We will construct a table that lists the derivatives of and the anti-derivatives of . We can then obtain the final answer by a series of multiplications and additions or subtractions, as indicated by the arrows in the table.

Thus,

Thus,

Thus,

Thus,

### 4 Inverse Functions

We will now look at some examples the involving inverse functions and . We do not (yet) know anti-derivatives for these functions, so we will have to let equal the inverse function.

\begin{align*} \int x^5\ln (x) \;dx &= \frac{x^6}{6}\ln (x) - \int \frac{x^6}{6}\cdot \frac{1}{x} \, dx\\ \text{Now, the integrand } & \text{ can be simplified by canceling $x$ allowing us to integrate}\\ &= \frac{x^6}{6}\ln (x) - \frac 16\int x^5 \, dx\\ &= \frac{x^6}{6}\ln (x) - \frac{x^6}{36} + C\\ \end{align*}

This method can be used to anti-differentiate as well.

Then and .

By IBP, \begin{align*} \int \tan ^{-1}(2x) dx &= x\tan ^{-1}(2x) - \int \frac{2x}{1+4x^2} \;dx\\ \text{we can use $u$-} & \text{substitution on this integral} \\ \text{with $u =$} & 1+4x^2 \text{ and } du = 8x \, dx \\ &= x\tan ^{-1}(2x) - \frac 14 \int \frac{1}{u} \; du\\ &= x\tan ^{-1}(2x) - \frac 14 \ln |u| + C\\ &= x\tan ^{-1}(2x) - \frac 14 \ln (1+4x^2) + C \end{align*}

Then and .

By IBP, \begin{align*} \int \tan ^{-1}(x) dx &= \frac 14 x^4\tan ^{-1}(x) - \frac 14\int \frac{x^4}{1+x^2} \;dx\\ & \text{now use polynomial long division to get} \\ &= \frac 14 x^4\tan ^{-1}(x) - \frac 14 \int x^2 - 1 + \frac{1}{1+x^2} \; dx\\ &= \frac 14 x^4\tan ^{-1}(x) - \frac 14\left (\frac{x^3}{3} - x + \tan ^{-1}(x)\right ) + C\\ &= \frac 14 x^4\tan ^{-1}(x) - \frac{x^3}{12} + \frac{x}{4} - \frac 14 \tan ^{-1}(x) + C \end{align*}

### 5 Special Examples

Then and .

Thus Let and .

Then and .

Thus

Solving algebraically for the integral gives:

Then and .

Thus Let and .

Then and .

Thus

Solving algebraically for the integral gives:

Our next example involves the function and arises in the calculation of arc length of the parabola .

At the end of our computation, we will need to recall the following integral

We will call the original integral and use IBP, arriving back at the original integral (as in the previous example). Then, Using the trig identity , gives

and so and finally,

We will call the original integral and use IBP, arriving back at the original integral (as in the previous example). Let Then, Using the trig identity , gives and so and finally,

### 6 REDUCTION OF POWERS

### 7 APPLICATIONS

We use the method of shells and the formula with We get which we compute using integration by parts with This leads to

We use the method of shells and the formula with We get which we compute using integration by parts with This leads to