Techniques of Integration

A. Integration by parts

Recall the rule (f(x)g(x))' = f'(x)g(x) + f(x)g'(x). Now suppose that we try to find the indefinite integral

$$\int h(x)dx,$$

where $h(x)$ can be written in the form f(x)g'(x). It may perhaps be easier to find

\int g(x)f'(x)\,dx

than to find

$$\int h(x)dx.$$

By linearity,

\int (f(x)g(x))'\,dx = \int f'(x)g(x)\,dx + \int f(x)g'(x)\,dx,

or

f(x)g(x) = \int f'(x)g(x)\,dx + \int h(x)\,dx,

i.e.,

\int h(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx.

Thus an indefinite integral of $h(x)$ is found by subtracting from $f(x)g(x)$ an indefinite integral of f'(x)g(x).

Examples

B. Integration by substitution

Certain integrals become much simpler if one variable is replaced by a new one. Here one uses in place of the old variable a new one which is, in the range of integration, a strictly increasing or decreasing differentiable function of the old one. If $u$ is the new variable and $x$ the old one, we then have u'(x) = \dfrac{1}{x'(u)}, from the differentiation of inverse functions.

Suppose we are to find an indefinite integral $\int f(x)dx$. That is, we are to find a function $F(x)$ such that F'(x) = f(x). It may be that $x$ is a strictly increasing or decreasing differentiable function of some new variable $u$, say $x=x(u)$, such that $f(x(u))$, which we shall call $g(u)$, is a simpler function to integrate. Then we might try to find a function $G(u)$ such that G'(u) = g(u). If $u(x)$ is the inverse function to $x(u)$, we might try $H(x)=G(u(x))$ as an answer to our original problem. But

H'(x) = G'(u(x)) \cdot u'(x) = g(u(x)) \cdot u'(x) = f(x) \cdot u'(x),

which differs from $f(x)$ by the factor u'(x) = \dfrac{1}{x'(u)}. Therefore let us try instead to find a function $G(u)$ such that G'(u) = g(u) \cdot x'(u). Then set $F(x)=G(u(x))$. We have then

F'(x) = G'(u(x)) \cdot u'(x) = g(u) \cdot x'(u) \cdot u'(x) = g(u) = f(x),

or $F(x)$ is a function of the desired type.

There is a formal device for remembering this calculation, as follows: in $\int f(x)dx$, write $dx={\displaystyle \frac{dx}{du}}\cdot du$. If $f(x)=g(u)$, then

$$\int f(x)dx=\int g(u)\cdot \frac{dx}{du}\cdot du.$$

Of course, the answer on the right is a function of $u$, for which we must then substitute in terms of $x$. Note that the integral on the right is just \int g(u)x'(u)\,du, or our $G(u)$. Several examples will illustrate the technique.

Examples

C. Partial fractions

Suppose we want to find an indefinite integral of ${\displaystyle \frac{f(x)}{g(x)}}$, where $f(x)$ and $g(x)$ are polynomials (such functions are called rational functions). We may assume that the degree of $f(x)$ (the highest power of $x$ appearing) is less than the degree of $g(x)$; for if not, we can divide $f(x)$ by $g(x)$ and get

$$\frac{f(x)}{g(x)}=P(x)+\frac{h(x)}{g(x)},$$

where $P(x)$ is a polynomial and $h(x)$, the remainder in the division of $f(x)$ by $g(x)$, is of degree less than that of $g(x)$. Suppose also that we can factor $g(x)$ into products of polynomials of the forms $(x-a)$ and $x^2+bx+c$, . (It is always possible to do this, although it is usually difficult to determine the $a$'s, $b$'s and $c$'s.) Once this factorization is accomplished, we begin a procedure which enables us to integrate ${\displaystyle \frac{f(x)}{g(x)}}$.

What we do is to separate the factors of the above forms in $g(x)$ and try to write ${\displaystyle \frac{f(x)}{g(x)}}$ as a sum of quotients of polynomials. The denominators will be the separate factors of $g(x)$ and the numerators will have degrees less than those of the denominators. If a factor $x-a$ or $x^2+bx+c$ divides $g(x)$ more than once, say $n$ times, we have to allow for $n$ terms, one having the factor as denominator to each of the powers $1,2,\dots ,n$. In each of these terms, it is still sufficient to assume that the numerator is a constant or of degree one (in case the denominator is a power of $x^2+bx+c$). For instance, if we had something of the form

$$\frac{d{x}^3+e{x}^2+fx+g}{(x^2+bx+c{)}^2}=\frac{d{x}^3+e{x}^2}{(x^2+bx+c{)}^2}+\frac{fx+g}{(x^2+bx+c{)}^2},$$

we could divide $d{x}^3+e{x}^2$ by $x^2+bx+c$ to get a quotient $P(x)$ of degree 1 and a remainder $R(x)$ of degree at most 1. Then

$$\frac{d{x}^3+e{x}^2}{(x^2+bx+c{)}^2}=\frac{P(x)}{x^2+bx+c}+\frac{R(x)}{(x^2+bx+c{)}^2},$$

so that the expression is reduced to the form which we shall consider.

Let $g(x)=(x-a_1{)}^{m_1}\cdots (x-a_r{)}^{m_r}(x^2+b_{1}x+c_1{)}^{n_1}\cdots (x^2+b_{s}x+c_s{)}^{n_s}$. Let $n=m_1+\cdots +m_r+2n_1+\cdots +2n_s$. Then $f(x)$ has degree less than $n$, and we can write $f(x)=d_{n-1}{x}^{n-1}+d_{n-2}{x}^{n-2}+\cdots +d_{1}x+d_0$, where some of the $d$'s may be 0. Now we try to write

\begin{aligned} \frac{f(x)}{g(x)} = &\frac{A_{1,1}}{x-a_1} + \frac{A_{1,2}}{(x-a_1)^2} + \dots + \frac{A_{1,m_1}}{(x-a_1)^{m_1}} \\ &+ \frac{A_{2,1}}{x-a_2} + \dots + \frac{A_{r,m_r}}{(x-a_r)^{m_r}}\\ &+ \frac{B_{1,1}x+C_{1,1}}{x^2+b_1x+c_1} + \dots + \frac{B_{1,n_1}x+C_{1,n_1}}{(x^2+b_1x+c_1)^{n_1}} \\ &+ \dots + \frac{B_{s,n_s}x+C_{s,n_s}}{(x^2+b_sx+c_s)^{n_s}}. \end{aligned}

Here the $A,B,C$ are unknowns, and there are

$$m_1+\cdots +m_r+2n_1+\cdots +2n_s=n$$

of them in all. If we put the right-hand side of the equation over a common denominator, this denominator is $g(x)$. Therefore the numerator must be $f(x)$; so the constant term of the numerator must be $d_0$, the coefficient of $x$ must be $d_1$, etc. This gives us $n$ equations in the $n$ unknowns $A,B,C$, which we solve. Substituting these solutions in the right-hand side of the equation, it remains only to integrate that expression term-wise. Now

\begin{aligned} &\int \frac{1}{x-a}\,dx = \log(x-a); \\ &\int \frac{1}{(x-a)^n}\,dx = \frac{-1}{n-1}\cdot\frac{1}{(x-a)^{n-1}}, \quad n>1. \end{aligned}

Since $x^2+bx+c={(x+{\displaystyle \frac{b}{2}})}^2+(c-{\displaystyle \frac{b^2}{4}})$, , we can substitute $x+{\displaystyle \frac{b}{2}}$ for $x$ and assume that $x^2+bx+c$ is simply of the form $x^2+c$, .

\begin{aligned} &\int \frac{x}{x^2+c}\,dx = \frac{1}{2}\log(x^2+c); \\ &\int \frac{1}{x^2+c}\,dx = \frac{1}{\sqrt{c}}\arctan\left(\frac{x}{\sqrt{c}}\right);\\ &\int \frac{x}{(x^2+c)^n}\,dx = \frac{-1}{2(n-1)}\cdot\frac{1}{(x^2+c)^{n-1}}, \quad n>1; \end{aligned}

and

$$\int \frac{1}{(x^2+c{)}^n}dx$$

can be handled by the substitution $x=\sqrt{c}\tan u$, or $u=\arctan \left({\displaystyle \frac{x}{\sqrt{c}}}\right).$

Examples

By a suitable substitution, one can reduce integrals of rational functions in $\sin x$ and $\cos x$ to the problem of integrating rational functions of a new variable $u$. The trick is to let $u=\tan {\displaystyle \frac{x}{2}}$. Then $x=2\arctan u$, and

x'(u) = \frac{2}{1+u^2}.

Also,

$$\sin x=2\sin \frac{x}{2}\cos \frac{x}{2}=\frac{2\tan \frac{x}{2}}{{\sec }^2\frac{x}{2}}=\frac{2u}{1+u^2},$$

and

\begin{aligned} \cos x &= 2\cos^2\frac{x}{2} - 1 \\ &= \frac{2}{1+\tan^2\frac{x}{2}} - 1 \\ &= \frac{2}{1+u^2} - 1 \\ &= \frac{1-u^2}{1+u^2}. \end{aligned}

Thus, for instance,

$$\int \frac{1}{\sin x+\cos x}dx=\int \frac{1}{{\displaystyle \frac{2u}{1+u^2}}+{\displaystyle \frac{1-u^2}{1+u^2}}}\cdot \frac{2}{1+u^2}du$$$$=\int \frac{2}{2u+1-u^2}du.$$

Now $2u+1-u^2=-(u-(1+\sqrt{2}))(u-(1-\sqrt{2}))$, and we can apply the technique of partial fractions to complete the work.

Exercises