Section 5.6

Integration by Parts

Goals

– Present the formula for integration by parts

– Apply the formula to finding various antiderivatives

Introduction

Every differentiation rule has a corresponding integration rule.

– For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation.

The rule that corresponds to the Product Rule for differentiation is called the rule for integration by parts.

Integration by Parts (cont’d)

The Product Rule says that if f and g are differentiable functions, then

In the notation for definite integrals this equation becomes

Integration by Parts (cont’d)

This is the same as

This leads to the formula for integration by parts:

Another Version

The formula may be easier to remember in the following notation.

Let u = f(x) and v = g(x). Then du = f (x) dx and dv = f (x) dx.

So by the Substitution Rule, the formula for integration by parts becomes

Example

Find ∫ x sin x dx .

Solution using first formula Suppose we choose f(x) = x and g (x) = sin x. Then f (x) = 1 and g(x) = – cos x.

– Note that for g we can use any antiderivative of g .

This leads to the calculations shown on the next slide:

Solution (cont’d)

Differentiating our answer (a wise practice!) confirms that it is correct.

Solution (cont’d)

Using the second formula Let u = x and dv = sin x dx, so du = dx and v = – cos x.

Remark

The goal of integration by parts is to get a simpler integral than the original one.

In the preceding example, we could have chosen u = sin x and dv = x dx; then

integration by parts would have led to

Remark (cont’d)

Whereas this a true statement, the integral on the right is a more difficult one than the one we started with!

In general, we try to choose…

– u = f(x) to be a function that becomes simpler when differentiated, as long as

– dv = g (x) dx can be integrated to give v.

Example

Evaluate ∫ ln x dx .

Solution Let u = ln x, so du = (1/x)dx, v = x.

Integrating by parts gives

Example

Find ∫ t 2 e t dt .

Solution Since t 2 becomes simpler (whereas e t is unchanged) when differentiated, we choose u = t 2 and dv = e t dt.

Then du = 2t dt and v = e t, and integration by parts gives

Solution (cont’d)

The new integral, ∫ t et dt, is simpler than the original but is still not obvious.

Using integration by parts again, this time with u = t and dv = dt. Then du = dt, v = et, and

Solution (cont’d)

Combining our results gives

Definite Integrals

Combining the formula for integration by parts with the Evaluation Theorem gives

As an example we calculate

1 1

0tan :xdx

Solution

We let u = tan-1x and dv = dx, so that

2 and

1

dxdu v x

x

Solution (cont’d)

To evaluate this integral we use t =1+x 2. Then dt = 2x dx, so x dx = dt/2.

When x = 0, t = 1; when x = 1, t = 2; so

Solution (cont’d)

Therefore

Review

Formula for integration by parts

Examples/strategies for using the formula

Application to definite integrals

Section 5.7

Additional Techniques of Integration

Goals

– Use trigonometric identities to integrate certain combinations of trigonometric functions

– Learn the methods of trigonometric substitution and partial fractions

Trigonometric Integrals

We can use trigonometric identities to integrate certain combinations of trigonometric functions.

As an example, we evaluate ∫ cos3 x dx .

Solution We separate one cosine factor and convert the remaining cos2 x factor using the identity sin2 x + cos2 x = 1 :

Solution (cont’d)

We can then evaluate the integral by substituting u = sin x, so du = cos x dx and

Example

Evaluate ∫ sin2 x dx .

Solution Because the half-angle formula sin2 x = ½(1 – cos 2x):

Trigonometric Substitution

A number of practical problems require us to integrate algebraic functions containing an expression of the form

or

Sometimes, the best way to perform the integration is to make a trigonometric substitution that gets rid of the integral sign.

2 2 2 2,a x a x2 2 .x a

Example

Show that area of a circle of radius r is πr 2.

Solution For simplicity we take the equation of the circle to be x2 + y2 = r2; solving for y gives

By symmetry the total area A is four times the area in the first quadrant, so

2 2y r x

2 2

0

1

4

rA r x dx

Solution (cont’d)

We want to make a substitution that turns r2 – x2 into the square of something.

Since

r2 – r2 sin2 θ = r2(1 – sin2 θ) = r2 cos2 θ,

we make the substitution x = rsin θ.

Because 0 ≤ x ≤ r, we restrict θ so that 0 ≤ θ ≤ π/2. We have dx = rcosθ dθ and

Solution (cont’d)

(Note that cosθ ≥ 0 when 0 ≤ θ ≤ π/2.)

Therefore, the Substitution Rule gives

In a manner similar to the preceding example, we integrate cos2 θ using the identity cos2 θ = ½(1 – cos 2θ):

Solution (cont’d)

This proves the famous formula for the area of a circle!

Partial Fractions

This method allows us to integrate rational functions.

Solution Notice that the denominator can be factored as a product of linear factors:

2

5 4:

2 1

xdx

x x

Solution (cont’d)

In a case like this, where the numerator has a smaller degree than the denominator, we can write the given rational function as a sum of partial fractions:

Solution (cont’d)

To find the values of the constants A and B, we multiply both sides of this equation by (x + 1)(2x – 1), obtaining

5x – 4 = A(2x – 1) + B(x + 1), or

5x – 4 = (2A + B)x + (–A + B)

The coefficients of x must be equal and the constant terms are also equal. So

Solution (cont’d)

2A + B = 5 and –A + B = –4

Solving these linear equations for A and B, we get A = 3 and B = –1, so

Finally, the resulting partial fractions are easy to integrate by substitution:

Solution (cont’d)

We offer some remarks about this method:

– If the degree in the numerator is the same as that of the denominator, or higher, we must first perform long division, for instance…

Remarks (cont’d)

– If the denominator has more than two linear factors, then we need to include a term corresponding to each factor. For example,

where A, B, and C are constants to be found.

Remarks (cont’d)

– If a linear factor is repeated, then we need to include extra terms in the partial fraction expansion, for example

– When we factor a denominator as far as possible, it might happen that we obtain an irreducible quadratic factor ax2 + bx + c, where the discriminant b2 – 4ac is negative. Then…

Remarks (cont’d)

– …the corresponding partial fraction is of the form

where A and B are constants to be determined. This term can be integrated by completing the square and using the formula

Review

Integrating trigonometric integrals

The method of trigonometric substitution

The method of partial fractions

Section 5.10

Improper Integrals

Goals

– Define improper integrals of Types 1 and 2

– Study convergence and divergence of improper integrals

– Use comparison to help decide whether an improper integral converges

Introduction

In defining a definite integral we assumed that…

– f is defined on a finite interval [a, b] and that

– f has no infinite discontinuity.

We want to extend this to the cases where – the interval is infinite, and/or

– f has an infinite discontinuity.

Such integrals are called improper.

b

af x dx

Type 1: Infinite Intervals

Let S be the infinite region that lies

– under the curve y = 1/x2,

– above the x-axis, and

– to the right of the

line x = 1.

Is the area of S infinite?

Type 1 (cont’d)

The area of the part of S that lies to the left of the line x = t is

Notice that A(t) < 1 no matter how large t is chosen, and A(t) 1 as t ∞.

Type 1 (cont’d)

So we say that the area of the infinite region S is equal to 1 and we write

With this example in mind we define the integral of f over an infinite interval as the limit of integrals over finite intervals:

Type 1 (cont’d)

Type 1 (cont’d)

For example,

11/ is divergent, sincex dx

Type 1 (cont’d)

Thus we have shown that

Type 1 (cont’d)

Example

Evaluate

Solution Part (b) of the definition gives

We integrate by parts with u = x, dv = ex dx, du = dx, v = ex :

0

.xxe dx

Solution (cont’d)

We know that et 0 as t –∞; so by l’Hospital’s

Rule we have

Therefore

Example

For what values of p is the integral

convergent?

Solution We know from our first example that the integral diverges if p = 1; for p ≠ 1,

1

1p dxx

Solution (cont’d)

If p > 1, then p – 1 > 0, so as t ∞, t p-1 ∞ and 1/t p-1 0. Therefore

and so the integral converges.

But if p < 1, then p – 1 < 0 and so…

Solution (cont’d)

and the integral diverges.

We summarize this result for future reference:

Type 2: Discontinuous Integrands

Suppose that f is

– a positive continuous function defined on a finite interval [a, b) but has

– a vertical asymptote at b.

Let S be the unbounded region under the graph of f and above the x-axis between a and b, as shown on the next slide:

Type 2 (cont’d)

Type 2 (cont’d)

The area of the part of S between a and t is

If it happens that A(t) approaches a number A as t b –, then we say that the

area of the region S is A and we write

t

aA t f x dx

limb t

a at bf x dx f x dx

Type 2 (cont’d)

We use this equation to define an improper integral of Type 2…

– even when f is not positive function,

– no matter what type of discontinuity f has at b.

The formal definition is found on the next slides:

Definition

Definition (cont’d)

Definition (cont’d)

Example

Find

Solution First, this integral is improper

because

asymptote x = 2.

Since the infinite discontinuity occurs at the left endpoint of [2, 5], we use part (b) of the definition:

5

2

1.

2dx

x

1/ 2 has the verticalf x x

Solution (cont’d)

Thus, the integral converges. Since the integrand is positive, we can interpret the value of the integral as the area shown on the next slide:

Solution (cont’d)

Example

Evaluate

Solution Note that the line x = 1 is a vertical asymptote of the integrand.

Since it occurs in the middle of [0, 3], we need part (c) of the definition with c = 1:

3

0 if possible.

1

dx

x

Solution (cont’d)

because 1 – t 0+ as t 1– .

Thus the integral diverges (even though

1

00 01 1

1

1

Now = lim limln 11 1

lim ln 1 ln 1

limln 1

t t

t t

t

t

dx dxx

x xt

t

3

1 converges).

1

dx

x

A Comparison Test

Sometimes it is impossible to find the exact value of an improper integral…

…and yet it is important to know whether it is convergent or divergent.

In such cases a comparison test such as the following theorem can be very useful.

We state the test for Type 1 integrals; a similar version holds for Type 2:

Comparison Test (cont’d)

The idea behind the theorem is that if

– the area under the graph of f is finite, then so is the area under the graph of g;

– the area under the graph of g is infinite, then so is the area under the graph of f.

Example

Show that

Solution It is not possible to evaluate the integral directly since the antiderivative of is not an elementary function

Instead we write

2

0 is convergent.xe

2

xe

Solution (cont’d)

and note that the first integral on the right is just an ordinary definite integral.

In the second integral we use the fact that for x ≥ 1 we have x2 ≥ x, so –x2 ≥ –x and

therefore

next slide.

The integral of e – x is easy to evaluate:

2

, as shown on thex xe e

Solution (cont’d)

Solution (cont’d)

Thus, taking f(x) = e – x and

in the Comparison Theorem shows that

It follows that

2xg x e

2

1 is convergent.xe dx

2

0 is convergent.xe dx

Review

Two types of improper integral

– Type 1 (infinite intervals)

– Type 2 (discontinuous integrands)

Comparison Theorem

Focus on Problem Solving

Evaluate

Solution We begin by looking at the “ingredients” of the function separately.

Here is how the first factor behaves when x approaches 3:

33

sinlim .

3

x

x

x tdt

x t

Solution (cont’d)

The second factor approaches

So as for the function overall…?

Now we recognize that…

– the Fundamental Theorem of Calculus could

be applied to

– using the definition of derivative:

3

3

sin0.

tdt

t

3

sinx tF x dt

t

Solution (cont’d)