# The Relation Between Integration and Differentiation - Part 2

#### 5.6 The Leibniz notation for primitives

We return now to a further study of the relationship between integration and differentiation. First we discuss some notation introduced by Leibniz.

We have defined a primitive $P$ of a function $f$ to be any function for which $P'(x) = f(x)$.
If is continuous on an interval, one primitive is given by a formula of the form

$P(x) = \int \limits_c^x f(t) \ dt$

and all other primitives cari differ from this one only by a constant. Leibniz used the symbol $\int f(x) dx$ to denote a general primitive off. In this notation, an equation like

(5.12) $\int f(x) \ dx = P(x) + c$

is considered to be merely an alternative way of writing $P'(x) = f(x)$.

For example, since the derivative of the sine is the cosine, we may write

(5.13) $\int \cos x \ dx = \sin x + C$

Similarly, since the derivative of $\frac{x^{n+1}}{n+1}$ is $x^n$, we may write

(5.14) $\int x^n \ dx = \frac{x^{n+1}}{n+1} + C$

for any rational power n ≠ - 1. The symbol C represents an arbitrary constant so each of Equations (5.13) and (5.14) is really a statement about a whole set of functions.

Despite similarity in appearance, the symbol $\int f(x) dx$ is conceptually distinct from the integration symbol $\int \limits_a^b f(x) \ dx$. The symbols originate from two entirely different processes-differentiation and integration. Since, however, the two processes are related by the fundamental theorems of calculus, there are corresponding relationships between the two symbols.

The first fundamental theorem states that any indefinite integral off is also a primitive of $f$. Therefore we may replace $P(x)$ in Equation (5.12) by $\int z f(t) dt$ for some lower limit $c$ and write (5.12) as follows:

(5.15) $\int f(x) \ dx = \int \limits_c^x f(t) \ dt + C$.

This means that we can think of the symbol $\int f(x) \ dx$ as representing some indefinite integral of $f$, plus a constant.

The second fundamental theorem tells us that for any primitive $P$ off and for any constant $C$, we have

$\int \limits_{a}^{b} f(x) \ dx = [P(x) + C]|_{a}^{b}$

If we replace $P(x) + C$ by $\int f(x) \ dx$, this formula may be written in the form

$\int \limits_a^b f(x) \ dx = \int f(x) \ dx |_a^b$

The two formulas in (5.15) and (5.16) may be thought of as symbolic expressions of the first and second fundamental theorems of calculus.

Because of long historical usage, many calculus textbooks refer to the symbol $\int f(x) \ dx$ as an "indefinite integral" rather than as a primitive or an anti-derivative. This is justified, in part, by Equation (5.15), which tells us that the symbol $\int f(x) dx$ is, apart from an additive constant $C$, an indefinite integral off. For the same reason, many handbooks of mathematical tables contain extensive lists of formulas labeled “tables of indefinite integrals” which, in reality, are tables of primitives. To distinguish the symbol $\int f(x) \ dx$ from $\int \limits_a^b f(x) \ dx$, the latter is called a dejnite integral. Since the second fundamental theorem reduces the problem of integration to that of finding a primitive, the term “technique of integration” is used to refer to any systematic method for finding primitives. This terminology is widely used in the mathematical literature, and it Will be adopted also in this book. Thus, for example, when one is asked to "integrate" $\int f(x) \ dx$, it is to be understood that what is wanted is the most general primitive off.

There are three principal techniques that are used to construct tables of indefinite integrals, and they should be learned by anyone who desires a good working knowledge of calculus. They are (1) integration by substitution (to be described in the next section), a method based on the chain rule; (2) integration byparts, a method based on the formula for differentia ting a product (to be described in Section 5.9); and (3) integration by-partial fractions. These techniques not only explain how tables of indefinite integrals are constructed, but also they tell us how certain formulas are converted to the basic forms listed in the tables.

#### 5.7 Integration by substitution

Let $Q$ be a composition of two functions $P$ and $g$, say $Q(x) = P[g(x)]$ for all $x$ in some interval $Z$.
If we know the derivative of $P$, say $P'(x) = f(x)$, the chain rule tells us that the derivative of $Q$ is given by the formula $Q'(x) = P'[g(x)]g'(x)$. Since $P' = f$, this states that $Q'(x) = f[g(x)]g'(x)$. In other words,

(5.17) $P'(x) = f(x)$ implies $Q'(x) = f[g(x)]g'(x)$.

In Leibniz notation, this statement cari be written as follows: If we have the integration formula

(5.18) $\int f(x) \ dx = P(x) + C$,

then we also have the more general formula

(5.19) $\int f[g(x)]g'(x) \ dx = P[g(x)] + C$.

For example, if $f(x) = cosx$, then (5.18) holds with $P(x) = sinx$, so (5.19) becomes

(5.20) $\int \cos g(x) . g'(x) dx = \sin g(x) + C$.

In particular, if g(x) = x^{3}, this gives us

cosx^{3}.3x^{2}dx = sinx^{3} + C,

a result that is easily verified directly since the derivative of sinx^{3} is 3x^{2}cosx^{3}.

Now we notice that the general formula in (5.19) is related to (5.18) by a simple mechanical process. Suppose we replace $g(x)$ everywhere in (5.19) by a new symbol u and replace $g'(x)$ by $\frac{du}{dx}$, the Leibniz notation for derivatives. Then (5.19) becomes

$\int f(u) \frac{du}{dx} \ dx = P(u) + C$

At this stage the temptation is strong to replace the combination $\frac{du}{dv}dx$ by $du$. If we do this, the last formula becomes

(5.21) $\int f(u) \ du = P(u) + C$

Notice that this has exactly the same form as (5.18), except that the symbol $u$ appears everywhere instead of $x$. In other words, every integration formula such as (5.18) can be made to yield a more general integration formula if we simply substitute symbols. We replace $x$ in (5.18) by a new symbol u to obtain (5.21), and then we think of u as representing a new function of $x$, say $u = g(x)$. Then we replace the symbol $du$ by the combination $g'(x) dx$, and Equation (5.21) reduces to the general formula in (5.19).

For example, if we replace $x$ by $u$ in the formula $\int \cos x \ dx = \sin x + C$, we obtain

$\int \cos u \ du = \sin u + C$

In this latter formula, $u$ may be replaced by $g(x)$ and $du$ by $g'(x)dx$, and a correct integration formula, (5.20), results.

When this mechanical process is used in reverse, it becomes the method of integration by substitution. The abject of the method is to transform an integral with a complicated integrand, such as $\int 3x^2 \cos x^3 \ dx$, into a more familiar integral, such as $\int \cos u \ du$. The method is applicable whenever the original integral cari be written in the form

$\int f[g(x)] g'(x) \ dx$

since the substitution

$u = g(x), du = g'(x)dx$,

transforms this to $\int f(u) \ du$. If we succeed in carrying out the integration indicated by $\int f(u) \ du$, we obtain a primitive, say $P(u)$, and then the original integral may be evaluated by replacing $u$ by $g(x)$ in the formula for $P(u)$.

The reader should realize that we have attached no meanings to the symbols $dx$ and $du$ by themselves. They are used as purely forma1 devices to help us perform the mathematical operations in a mechanical way. Each time we use the process, we are really applying the statement (5.17).

Success in this method depends on one's ability to determine at the outset which part of the integrand should be replaced by the symbol $u$, and this ability cornes from a lot of experience in working out specific examples. The following examples illustrate how the method is carried out in actual practice.

**EXAMPLE 1.** Integrate $\int x^3 \cos x^4 \ dx$.

Solution. Let us keep in mind that we are trying to write $x^3cos(x^4)$ in the form $f[g(x)]g'(x)$ with a suitable choice off and $g$. Since cosx^{4} is a composition, this suggests that we take $f(x) = cosx$ and $g(x) = x^4$ so that $cosx^4$ becomes $f[g(x)]$. This choice of $g$ gives $g'(x) = 4x^3$, and hence f[g(x)]g'(x) = (cosx^{4}) (4x^{3}). The extra factor 4 is easily taken care of by multiplying and dividing the integrand by 4. Thus we have

x^{3}cosx^{4} = (cosx^{4})(4x^{3})/4 = {f[g(x)]g'(x)}/4.

Now, we make the substitution u = g(x) = x^{4}, du = g'(x)dx = 4x^{3}dx, and obtain

$\int x^3 \cos x^4 \ dx = \frac{1}{4} \int f(u) \ du = \frac{1}{4} \int \cos u \ du = \frac{1}{4} \sin u + C$

Replacing $u$ by $x^4$ in the end result, we obtain the formula

$\int x^3 \cos x^4 \ dx = \frac{1}{4} \sin x^4 + C$

which can be verified directly by differentiation.

After a little practice one carry perform some of the above steps mentally, and the entire calculation can be given more briefly as follows: Let u = x^{4}. Then du = 4x^{3}dx, and we obtain

$\int x^3 \cos x^4 \ dx = \frac{1}{4} \int (\cos x^4) (4x^3 \ dx) = \frac{1}{4} \int \cos u \ du = \frac{1}{4} \sin u + C = \frac{1}{4} \sin x^4 + C$

Notice that the method works in this example because the factor $x^3$ has an exponent one less than the power of $x$ which appears in $cos(x^4)$.

**EXAMPLE 2.** Integrate $\int \cos^2 x \ \sin x \ dx$.

Solution. Let $u = cos x$. Then $du = -sin x dx$, and we get

$\int \cos^2 x \ \sin x \ dx = - \int (\cos x)^2 (-\sin x \ dx) = - \int u^2 \ du = - \int u^2 \ du = - \frac{u^3}{3} + C = - \frac{\cos^3 x}{3} + C$

Again, the final result is easily verified by differentiation.

EXAMPLE 3. Evaluate $\int \limits_2^3 \frac{(x+1)\ dx}{\sqrt{x^2 + 2x +3}}$

Solution. Let u = x^{2} + 2x + 3. Then du = (2x + 2)dx, so that

$\frac{(x+1)\ dx}{\sqrt{x^2 + 2x +3}} = \frac{1}{2} \frac{du}{\sqrt{u}}$

Now we obtain new limits of integration by noting that u = 11 when x = 2, and that u = 18 when x = 3. Then we write

$\int \limits_2^3 \frac{(x+1)\ dx}{\sqrt{x^2 + 2x +3}} = \frac{1}{2} \int \limits_{11}^{18} u^{-\frac{1}{2}} \ du = \sqrt{u}|_{11}^{18} = \sqrt{18} - \sqrt{11}$. The same result is arrived at by expressing everything in terms of $x$. Now we prove a general theorem which justifies the process used in Example 5.

**THEOREM 5.4. SUBSTITUTION THEOREM FOR INTEGRALS.** Assume $g$ has a continuous derivative $g'$ on an open interval $I$.
Let $J$ be the set of values taken by $g$ on $I$ and assume that $f$ is continuous on $J$. Then for each $x$ and $c$ in $I$, we have

(5.22) $\int \limits_c^x f[g(t)] g'(t) \ dt = \int \limits_{g(c)}^{g(x)} f(u) \ du$

**Proof.** Let $a = g(c)$ and define two new functions $P$ and $Q$ as follows:

$P(x) = \int \limits_a^x f(u) \ du$ if $x$ is from $J$

$Q(x) = \int \limits_c^x f[g(t)] g'(t) \ dt$ if $x$ is from $I$.

Since $P$ and $Q$ are indefinite integrals of continuous functions, they have derivatives given by the formulas

P'(x) = f(x), Q'(x) = f[g(x)]g'(x).

Now let R denote the composite function, R(x) = P[g(x)]. Using the chain rule, we find

R'(x) = P'[g(x)]g'(x) = fg(x)g'(x) = Q'(x).

Applying the second fundamental theorem twice, we obtain

$\int \limits_{g(c)}^{g(x)} f(u) \ du = \int \limits_{g(c)}^{g(x)} P'(u) \ du = P[g(x)] - P[g(c)] = R(x) - R(c)$

and

$\int \limits_c^x f[g(t)] g'(t) \ dt = \int \limits_c^x Q'(t) \ dt = \int \limits_c^x R'(t) \ dt = R(x) - R(c)$

This shows that the two integrals in (5.22) are equal.

**5.8 Exercises**

In Exercises 1 through 20, evaluate the integrals by the method of substitution.

- $\int \sqrt{2x+1} \ dx$
- $\int x \sqrt{1+3x} \ dx$
- $\int x^2 \sqrt{x+1} \ dx$
- $\int \limits_{-\frac{2}{3}}^{\frac{1}{3}} \frac{x \ dx}{\sqrt{2 - 3x}}$
- $\int \frac{(x+1) \ dx}{(x^2+2x+2)^3}$
- $\int \sin^3 x \ dx$
- $\int z(z-1)^{\frac{1}{3}} \ dz$
- $\int \frac{\cos x \ dx}{\sin^3 x}$
- $\int \limits_0^{\frac{\pi}{4}} \cos 2x\sqrt{4-\sin 2x} \ dx$
- $\int \frac{\sin x \ dx}{(3 + \cos x)^2}$
- $\int \frac{\sin x \ dx}{\sqrt{\cos^3 x}}$
- $\int \limits_3^8 \frac{\sin \sqrt{x+1} \ dx}{\sqrt{x+1}}$
- $\int x^{n-1} \sin x^n \ dx \qquad \qquad n\neq 0$
- $\int \frac{x^5 \ dx}{\sqrt{1-x^6}}$
- $\int t(1+t)^{\frac{1}{4}} \ dt$
- $\int (x^2 + 1)^{-\frac{3}{2}} \ dx$
- $\int x^2(8x^3 + 27)^{\frac{2}{3}} \ dx$
- $\int \frac{(\sin x + \cos x) \ dx}{(\sin x - \cos x)^{\frac{1}{3}}}$
- $\int \frac{x \ dx}{\sqrt{1+x^2+\sqrt{(1+x^2)^3}}}$
- $\int \frac{(x^2 + 1 - 2x)^{\frac{1}{5}} \ dx}{1-x}$

21. Deduce the formulas in Theorems 1.18 and 1.19 by the method of substitution.

22. Let

$F(x,a) = \int \limits_0^x \frac{t^p}{(t^2+a^2)^q} dt$

where a > 0, and p and q are positive integers. Show that F(x, a) = a^{p + 1 - 2q}F(x/a, 1).

23. If m and n are positive integers, show that

$\int \limits_0^1 x^m (1-x)^n \ dx = \int \limits_0^1 x^n(1-x)^m \ dx$

24. Show that $\int \limits_0^\pi xf(\sin x) \ dx = \frac{\pi}{2} \int \limits_0^\pi f(\sin x) \ dx \qquad \qquad [Hint: u=\pi - x]$

#### 5.9 Integration by parts

We proved that the derivative of a product of two functions $f$ and $g$ is given by the formula

$h'(x) = f(x)g'(x) + f'(x)g(x)$,

where $h(x) =f(x)g(x)$. When this is translated into the Leibniz notation for primitives, it becomes $\int f(x)g'(x) \ dx + \int f'(x)g(x) \ dx = f(x)g(x) + C$, usually written as follows:

(5.23) $\int f(x)g'(x) \ dx = f(x)g(x) - \int f'(x)g(x) \ dx + C$.

This equation, known as the formula for integration by parts, provides us with a new integration technique.

To evaluate an integral, say $\int k(x) \ dx$, using (5.23), we try to find two functions $f$ and $g$ such that $k(x)$ can be written in the form $f(x)g'(x)$. If we carry do this, then (5.23) tells us that we have

$\int k(x)dx = f(x)g(x) - \int g(x)f'(x)dx + C$, and the difficulty has been transferred to the evaluation of $\int g(x) f'(x) \ dx$. If $f$ and $g$ are properly chosen, this last integral may be easier to evaluate than the original one. Sometimes two or more applications of (5.23) Will lead to an integral that is easily evaluated or that may be found in a table. The examples worked out below have been chosen to illustrate the advantages of this method. For definite integrals, (5.23) leads to the formula

$\int \limits_a^b f(x) g'(x) \ dx = f(b)g(b) - f(a)g(a) - \int \limits_a^b f'(x) g(x) \ dx$

If we introduce the substitutions $u = f(x), u = g(x), du =f'(x) dx$, and $dv = g'(x) dx$, the formula for integration by parts assumes an abbreviated form that many people find easier to remember, namely

(5.24) $\int u \ dv = uv - \int v \ du + C$

**EXAMPLE 1.** Integrate $\int x \cos x \ dx$

Solution. We choose $f(x) = x$ and $g'(x) = cos x$. This means that we have $f'(x) = 1$ and $g(x) = sin x$, so (5.23) becomes

(5.25) $\int x \cos x \ dx = x \sin x - \int \sin x \ dx + C = x \sin x + \cos x +C$.

Note that in this case the second integral is one we have already calculated.

TO carry out the same calculation in the abbreviated notation of (5.24) we write

u = x, dv = cosxdx,

du = dx, v = _{j}cosxdx = sinx,

$\int x \cos x \ dx = uv - \int v \ du = x \sin x - \int \sin x \ dx + C = x \sin x + \cos x +C$

Had we chosen $u = cos x$ and $du = x dx$, we would have obtained $du = -sin x dx$, v = x^{2}/2, and (5.24) would have given us

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

Since the last integral is one which we have not yet calculated, this choice of u and u is not as useful as the first choice. Notice, however, that we cari salve this last equation for $\int x^2 \sin x \ dx$ and use (5.25) to obtain

$\int x^2 \sin x \ dx = 2x \sin x + 2 \cos x - x^2 \cos x + C$

As an application of the method of integration by parts, we obtain another version of the weighted mean-value theorem for integrals (Theorem 3.16).

**THEOREM 5.5. SECOND MEAN-VALUE THEOREM FOR INTEGRAL.** Assume $g$ is continuous on $[a, b]$, and assume $f$ has a derivative which is continuous and never changes sign in [a, b].Then, for some c in [a, b], we have

(5.28) $\int \limits_a^b f(x) g(x) \ dx = f(a) \int \limits_a^c g(x) \ dx + f(b) \int \limits_c^b g(x) \ dx$

**Proof.** Let $G(x) = \int \limits_a^x g(t) \ dt$. Since g is continuous, we have $G'(x) = g(x)$. Therefore, integration by parts gives us

(5.29) $\int \limits_a^b f(x) g(x) \ dx = \int \limits_a^b f(x) G'(x) \ dx = f(b) G(b) - \int \limits_a^b f'(x) G(x) \ dx$

since G(a) = 0. By the weighted mean-value theorem, we have

$\int \limits_a^b f'(x) G(x) \ dx = G(c) \int \limits_a^b f'(x) \ dx = G(c) [f(b) - f(a)]$

for some $c$ in $[a, b]$. Therefore, (5.29) becomes

$\int \limits_a^b f(x) g(x) \ dx = f(b)G(b) - G(c)[f(b) - f(a)] = f(a)G(c) + f(b)[G(b) - G(c)]$

This proves (5.28) since $G(c) = \int \limits_a^c g(x) \ dx \qquad and \qquad G(b) - G(c) = \int \limits_c^b g(x) \ dx$.

#### 5.10 Exercises

Use integration by parts to evaluate the integrals in Exercises 1 through 6.

- $\int x \sin x \ dx$
- $\int x^2 \sin x \ dx$
- $\int x^3 \cos x \ dx$

- $\int x^3 \sin x \ dx$
- $\int \sin x \cos x \ dx$
- $\int x \sin x \cos x \ dx$

$\int \sin^2 x \ dx = - \sin x \cos x + \int \cos^2 x \ dx$

In the second integral, write cos

^{2}x = 1 - sin

^{2}x and thereby deduce the formula

$\int \sin^2 x \ dx = \frac{1}{2} x - \frac{1}{4} \sin 2x$

7. Derive the following formulas.

- $\int \sin^3 x \ dx = -\frac{3}{4} \cos x + \frac{1}{12} \cos 3x$
- $\int \sin^4 x \ dx = -\frac{3}{8} x - \frac{1}{4} \sin 2x + \frac{1}{32} \sin 4x$
- $\int \sin^5 x \ dx = -\frac{5}{8} x + \frac{5}{48} \cos 3x - \frac{1}{80} \cos 5x$

- $\int x \sin^2 x \ dx = \frac{1}{4} x^2 - \frac{1}{4} x \sin 2x - \frac{1}{8} \cos 2x$
- $\int x \sin^3 x \ dx = \frac{3}{4} \sin x - \frac{1}{36} \sin 3x - \frac{3}{4} x \cos x + \frac{1}{12} x \cos 3x$
- $\int x^2 \sin^2 x \ dx = \frac{1}{6} x^3 + ( \frac{1}{8} - \frac{1}{4} x^2) \sin 2x - \frac{1}{4} x \cos 2x$

$\int \sqrt{1 - x^2} \ dx = x \sqrt{1 - x^2} + \int \frac{x^2}{\sqrt{1 - x^2}} \ dx$

Write x

^{2}= x

^{2}- 1 + 1 in the second integral and deduce the formula

$\int \sqrt{1 - x^2} \ dx = \frac{1}{2} x \sqrt{1 - x^2} + \frac{1}{2} \int \frac{1}{\sqrt{1 - x^2}} \ dx$

10. If $I_n (x) = \int \limits_0^x t^n (t^2 + a^2)^{-\frac{1}{2}} \ dt$, use integration by parts to show that

$n I_n (x) = x^{n-1} \sqrt{x^2 + a^2} - (n-1) a^2 I_{n-2} (x) \qquad \qquad if \ n \geq 2$

11. Evaluate the integral $\int \limits_{-1}^3 t^3 (4 + t^3)^{-\frac{1}{2}} \ dt$, given that $\int \limits_{-1}^3 (4 + t^3)^{\frac{1}{2}} \ dt = 11.35$. Leave the answer in terms of $\sqrt{3}$ and $\sqrt{31}$.

#### *5.11 Miscellaneous review exercises

1. Let $f$ be a polynomial with $f(0) = 1$ and let $g(x) = PJ'(x)$. Compute $g(0), $g'(0),...,g^{(n)}(0)$.

2. Find a polynomial P of degree ≤ 5 with P(0) = 1,P(1) = 2,P'(0) = P''(0) = P'(1) = P''(1) = 0.

3. If f(x) = cosx and g(x) = sinx, prove that

,f^{(n)}(x) = cos(x + nπ/2) and g^{(n)}(x) = sin(x + nπ/2).

4. If h(x) = f(x)g(x), prove that the nth derivative of h is given by the formula

$h^{(n)} (x) = \sum\limits_{k=0}^n {{n} \choose {k}} f^{(k)} (x) g^{(n-k)} (x)$

This is called Leibniz's formula.

5. Given two functions f and g whose derivatives f' and g' satisfy the equations

(5.30) f'(x) = g(x), g'(x) = -f(x), f(0) = 0, g(0) = 1,

for every x in some open interval J containing 0. For example, these equations are satisfied when f (x) = sinx and g(x) = cosx.

(a) Prove that f^{2}(x) + g^{2}(x) = 1 for every x in J.

(b) Let F and G be another pair of functions satisfying (5.30). Prove that F(x) =f(x) and G(x) = g(x) for every x in J. [Hint: Consider h(x) = [F(x) -,f(x>]^{2} + [G(x) - g(x)]^{2}.]

(c) What more cari you say about functionsfand g satisfying (5.30)?

6. A function f, defined for all positive real numbers, satisfies the equation f(x^{2}) = x^{3} for every x > 0. Determine f'(4).

7. A function g, defined for all positive real numbers, satisfies the following two conditions: g(1) = 1 and g'(x^{2}) = x^{3} for all x > 0. Compute g(4).

8. Let C_{1} and C_{2}, be two curves passing through the origin as indicated in Figure 5.2. A curve C is said to “bisect in area” the region between C_{1} and C_{2} if, for each point P of C, the two shaded regions A and B shown in the figure have equal areas. Determine the Upper curve C_{2},given that the bisecting curve C has the equation y = x^{2} and that the lower curve C_{1} has the equation y = x^{2}/2.

In Exercises 11 through 20, evaluate the given integrals. Try to simplify the calculations by using the method of substitution and/or integration by parts whenever possible.

- $\int (2 + 3x) \sin 5x \ dx$
- $\int x \sqrt{1 + x^2} \ dx$
- $\int\limits_{-2}^1 x (x^2 - 1)^9 \ dx$
- $\int\limits_0^1 \frac{2x + 3}{(6x + 7)^3} \ dx$
- $\int x^4 (1 + x^5)^5 \ dx$
- $\int \limits_0^1 x^4 (1 - x)^{20} \ dx$
- $\int \limits_1^2 x^{-2} \sin \frac{1}{x} \ dx$
- $\int \sin \sqrt[4]{x-1} \ dx$
- $\int x \sin x^2 \cos x^2 \ dx$
- $\int \sqrt{1 + 3 \cos^2 x} \sin 2x \ dx$

22. Determine a pair of numbers a and b for which $\int \limits_0^1 (ax + b) (x^2 + 3x + 2)^{-2} \ dx = \frac{3}{2}$

23. Let $F(m, n) = \int \limits_0^x t^m (1 + t^n) \ dt, m > 0, n > 0$. Show that

(m + 1)F(m, n) + nF(m + 1, n - 1) = x

^{m + 1}(1 + x)

^{n}.

Use this to evaluate F(10, 2).

24. Let $A$ denote the value of the integral

$\int \limits_0^\pi \frac{\cos x}{(x + 2)^2} \ dx$

Compute the following integral in terms of $A$:

$\int \limits_0^{\frac{\pi}{2}} \frac{\sin x \cos x}{x + 1} \ dx$

The formulas in Exercises 28 through 33 appear in integral tables. Verify each of these formulas by any method.

- $\int \frac{\sqrt{a + bx}}{x} \ dx = 2 \sqrt{a + bx} + a \int \frac{dx}{x \sqrt{a + bx}} + C " />
- $\int x^n \sqrt{ax + b} \ dx = \frac{2}{a (2n + 3)} \left( x^n (ax + b)^{\frac{3}{2}} - nb \int x^{n-1} \sqrt{ax + b} \ dx \right) + C \qquad ( n \neq -\frac{3}{2})$
- $\int \frac{x^m}{\sqrt{a + bx}} \ dx = \frac{2}{ (2m + 1)b} \left( x^m \sqrt{a + bx} - ma \int \frac{x^{m-1}}{ \sqrt{a + bx}} \ dx \right) + C \qquad ( m \neq -\frac{1}{2})$
- $\int \frac{dx}{x^n \sqrt{ax + b}} = - \frac{\sqrt{ax + b}}{(n-1) b x^{n-1}} - \frac{(2n - 3)a}{(2n - 2)b} \int \frac{dx}{x^{n-1} \sqrt{ax + b}} + C \qquad (n \neq 1)$
- $\int \frac{\cos^m x}{\sin^n x} \ dx = \frac{\cos^{m-1} x}{(m - n) \sin^{n-1} x} + \frac{m - 1}{m - n} \int \frac{\cos^{m-2} x}{\sin^n x} \ dx + C \qquad (m \neq n)$
- $\int \frac{\cos^m x}{\sin^n x} \ dx = \frac{\cos^{m+1} x}{(n - 1) \sin^{n-1} x} - \frac{m - n + 2}{n - 1} \int \frac{\cos^m x}{\sin^{n-2} x} \ dx + C \qquad (n \neq 1)$

^{2}. Prove that there is only one solution.

(b) If Q(x) is a given polynomial, prove that there is one and only one polynomial P(x) such that P'(x) - 3P(x) = Q(x).

35. A sequence of polynomials (called the Bernoullipolynomials) is defined inductively as follows:

P,(x) = 1; P'

_{n}(x) = nP

_{n - 1}(x) and $\int \limits_0^1 P_n (x) \ dx = 0$ if n ≥ 1.

(a) Determine explicit formulas for P

_{1}(x), P

_{2}(x), . . . , P

_{5}(x).

(b) Prove, by induction, that P

_{n}(x) is a polynomial in x of degree n, the term of highest degree being x

^{n}.

(c) Prove that P

_{n}(O) = P

_{n}(1) if n ≥ 2.

(d) Prove that P

_{n}(x + 1) - P

_{n}(x) = nx

^{n - 1}if n ≥ 1.

(e) Prove that for n ≥ 2 we have

$\sum \limits_{r=1}^{k-1} r^n = \int \limits_0^k P_n (x) \ dx = \frac{P_{n+1}(k) - P_{n+1}(0)}{n + 1}$

(f) Prove that P

_{n}(1 - x) = (-1)

^{n}P

_{n}(x) if n ≥ 1.

(g) Prove that P

_{2n + 1}(0) = 0 and P

_{2n - 1}(1/2) = 0 if n ≥ 1.

36. Assume that |f''(x)| ≤ m for each x in the interval [0, a], and assume that f takes on its largest value at an interior point of this interval. Show that |f'(0)| + |f'(a)| ≤ am. You may assume that f'' is continuous in [0, a].