Methods of Integration

References are to Thomas & Finney, 8th edition.

Integration The definition of the indefinite integral is

∫ du = u + C  where C is an arbitrary constant  (1)

for any variable u. The workhorse of integration is the method of substitution (or change of variable); see the flowchart [p. 517]. Integration is linear:

∫ (f(x) + g(x))dx = ∫ f(x)dx + ∫ g(x)dx  (2)

and

∫ kf(x)dx = k ∫ f(x)dx where k is constant.  (3)

Polynomials For polynomials we use

∫ u^{n}du = \frac{u^{n+1}}{n + 1} + C  provided n ≠ −1  (4)

where n may also be fractional or negative. The special case n = −1 is handled by

∫ \frac{du}{u} = log u + C  or log(−u) if u is negative.  (5)

Rational functions, I [See §7.5, esp. pp. 510-511.] Rational functions \frac{p(x)}{q(x)}, where

p(x) and q(x) are polynomials, can always be integrated.

If this fraction is improper, i. e. deg p(x) ≥ deg q(x), we must first use polynomial division

p(x) = q(x)s(x) + r(x),

where deg r(x) < deg q(x), to write it in terms of a proper fraction as

\frac{p(x)}{q(x)} = s(x) + \frac{r(x)}{q(x)}

We assume for now that q(x) is a product of linear factors. If the a_{i} are all distinct, any proper fraction can be decomposed uniquely as

\frac{p(x)}{(x − a_{1})(x − a_{2}) … (x − a_{n})} = \frac{A_{1}}{x − a_{1}} + \frac{A_{2}}{x − a_{2}} + … + \frac{A_{n}}{x − a_{n}}  (6)

for suitable constants A_{i}. This is the method of partial fractions. To find the A_{i}, clear the denominators and equate coefficients of powers of x, or choose various values for x, to obtain enough equations to determine the A_{i}. (Here, putting x = a_{i} is especially useful.) Then equation (6) is easily integrated by using equations (2) and (5) with the substitutions u = x − a_{i} and du = dx.

If the a_{i} are not all distinct, equation (6) is clearly inappropriate because the common denominator is wrong. In general, if q(x) has the repeated linear factor

(x − a)^{m}, we must replace the m identical terms \frac{A}{x − a} in equation (6) by

\frac{B_{1}}{x − a} + \frac{B_{2}}{(x − a)^{2}} + … + \frac{B_{m}}{(x − a)^{m}} .  (7)

This is easily integrated by equations (5) and (4).

Exponential functions These are handled by

∫ e^{u}du = e^{u} + C  (8)

Trigonometric functions The six trigonometric functions of x may be expressed in terms of cos x and sin x, so that the basic trigonometric polynomial integral takes the form ∫ sin^{m} x cos^{n} xdx. We can also allow m or n to be negative.

Case m odd We put u = cos x and du = − sin xdx and use sin^{2} x = 1 − u^{2} on the remaining even powers of sin x, to get a rational function of u.

Case n odd We put u = sin x and du = cos xdx and use cos^{2} x = 1 − u^{2} on the remaining even powers of cos x, to get a rational function of u.

Example One important and useful application is the integral

∫ sec xdx = ∫ \frac{cos xdx}{cos^{2} x} = ∫ \frac{du}{1 − u^{2}} = … = log(sec x + tan x) + C  (9)

Here, we use partial fractions to write

\frac{1}{1 − u^{2}} = \frac{1}{(1 + u)(1 − u)} = \frac{1/2}{1 + u} + \frac{1/2}{1 − u} .

By equation (5), this integrates to give

\frac{1}{2} log(1 + u) − \frac{1}{2} log(1 − u) = \frac{1}{2} log \frac{1 + sin x}{1 − sin x}

To clean this up, we write

\frac{1 + sin x}{1 − sin x} = \frac{(1 + sin x)(1 + sin x)}{(1 − sin x)(1 + sin x)} = \frac{(1 + sin x)^{2}}{1 − sin^{2} x}

= \frac{(1 + sin x)^{2}}{cos^{2} x} = (\frac{1 + sin x}{cos x})^{2} = (sec x + tan x)^{2}

and use log z^{2} = 2 log z.

Similarly, or by putting y = π/2 − x in equation (9), we have

∫ csc ydy = − log(csc y + cot y) + C  (10)

Case m and n even In this case we can use the double angle formulae

cos^{2} x = \frac{1 + cos 2x}{2} sin^{2} x = \frac{1 − cos 2x}{2}

to obtain an integral involving only cos 2x. Repeat if necessary.

If n is negative, the substitution u = tan x, du = sec^{2} xdx can be useful.

For integrals of the form ∫ sin mx sin nxdx etc., see p. 497.

Rational functions, II Not all polynomials have linear factors. However, we do have the fundamental theorem of real algebra:

Theorem 11 Every polynomial x^{n} + … in x factors uniquely up to order as a product of:

(i) linear factors of the form x − a; (ii) quadratic factors of the form x^{2} + ax + b that have no real root.

When q(x) has a quadratic factor x^{2} + ax + b, the appropriate term of the partial fraction decomposition must be taken as

\frac{Ax + B}{x^{2} + ax + b}  (12)

in order to provide enough indeterminates. For a repeated quadratic factor (x^{2} +ax+ b)^{m}, we need instead

\frac{A_{1}x + B_{1}}{x^{2} + ax + b} + \frac{A_{2}x + B_{2}}{(x^{2} + ax + b)^{2}} + … + \frac{A_{m}x + B_{m}}{(x^{2} + ax + b)^{m}}  (13)

Quadratic denominators (See §7.4.) First complete the square, if necessary, x^{2} + ax + b = (x + c)^{2} + f^{2}

where c = a/2 and f = \sqrt{b − a^{2}/4}, and make the linear substitution u = x + c and du = dx.

We break up equation (12) into two terms. In the first, the substitution u = f tan θ, du = f sec^{2} θdθ gives

∫ \frac{du}{u^{2} + f^{2}} = ∫ \frac{f sec^{2} θdθ}{f^{2} sec^{2} θ} = \frac{1}{f} θ + C = \frac{1}{f} tan^{−1} \frac{u}{f} + C  (14)

In the second, we simply put s = u^{2} + f^{2}, ds = 2udu, to get

∫ \frac{udu}{u^{2} + f^{2}} = ∫ \frac{ds}{2s} = \frac{1}{2} log s + C = \frac{1}{2} log(u^{2} + f^{2}) + C  (15)

However, the substitution u = f tan θ works here too, somewhat less efficiently. The same substitutions also handle the integrals

∫ \frac{du}{(u^{2} + f^{2})^{m}}  ∫ \frac{udu}{(u^{2} + f^{2})^{m}}

with repeated quadratic factors.