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INTEGRATION BY SUBSTITUTION

Integration by substitution. A very useful device for transforming a given integrand into a form in which the fundamental integration formulas can be applied is that of substituting a new variable. Consider the following example.

Example. Find

Solution. This integration cannot be performed directly by any of the standard formulas. Let us make the substitution

x = 2 sin θ

which will, we shall see, have the effect of transforming the integrand into a rational trigonometric function.

If x = 2 sin θ then dx = 2 cos θ dθ. We then have, upon substituting,

Now using the trigonometric identity cos 2θ = 2cos2 θ - 1 we get

= 2θ + sin 2θ + C .

We have thus performed the integration in terms of θ. We can now change the result back into terms of x as follows:

Since x = 2 sin θ,

Using this relationship we construct the triangle shown in Fig. 1. We know two sides of the triangle and get the other side from the Pythagorean theorem i.e. x2 + y2 = 22 so .

The trigonometric formula for sin 2θ is

sin 2θ = 2 sin θ cos θ .

Using Fig. 1 we have

Thus

Trigonometric substitutions. Integrands of the following types are often encountered in applications:

where m is a positive or negative integer. Integrands which are algebraic and involve no irrational function of x aside from integral powers of one of the quantities

can be transformed into rational trigonometric functions by a trigonometric substitution of the following type:

It is easily seen that these substitutions convert the integrands to rational expressions in terms of trigonometric functions since

Appropriate triangle constructions such as those shown in Fig. 2 obviate the use of trigonometric identities in transforming integrands.

Integrands containing quantities of types

can be dealt with in a similar way. In this case make the substitutions as follows:

Miscellaneous substitutions.

I If an integrand is rational except for a radical of the form

it can be rationalized by a substitution as follows:

substitute q + px - x2 = (α + x)2z2 or q + px - x2 = (β - x)2z2

II Any rational function of sin θ and cos θ can be replaced with a rational function of z by using the substitution

θ = 2 tan-1 z

since

The first and second of these relationships are obtained from Fig. 3 and the third by differentiating

θ = 2 tan-1 z .

The relationships shown in Fig. 3 are obtained from the trigonometric double-angle formula

which is equivalent to

Since θ = 2 tan-1 z ,

and substituting into 2) we obtain

from which we construct the triangle of Fig. 3.

III Theorem. If in

m, n, p, q are integers, the substitution

will rationalize

provided that

is a positive or negative integer or zero.

Evaluating integrals of types:

By using the process of “completing the square”, one can write any quadratic function Ax2 + Bx + C in one of the forms (u2 + a2), (u2 - a2), (a2 - u2), (-a2 - u2), possibly multiplied by a constant, where u is some linear function of x. The integrals 1) are thus turned into standard forms.

Reduction of Ax2 + Bx + C to one of the forms (u2 + a2), (u2 - a2), (a2 - u2) or (-a2 - u2). The function Ax2 + Bx + C can be written as

where p =B/A and q = C/A. Now we complete the square on x2 + px + q:

Thus we have

Set

and

From 2) we see that the function Ax2 + Bx + C can now be written as one of the forms (u2 + a2), (u2 - a2), (a2 - u2) or (-a2 - u2), depending on the sign of A (with the sign of A transferring).

Other substitutions.

1. The substitution ex = z reduces

2. The substitution tan x/2 = z reduces

3. The substitution x = 1/z reduces

Change of limits corresponding to a change of variable. When one does a change of variable in a definite integral he can avoid the trouble of changing the result back into terms of the original variable by making a corresponding change in the limits as illustrated in the following example.

Example. Evaluate

Solution. Let us use the substitution x = 4 sin θ. The integration is over the x-interval from x = 0 to x = 4. The corresponding interval for the new variable is found from the relation x = 4 sin θ as follows:

Putting x = 4:

4 = 4 sin θ,      sin θ = 1,         θ = π/2

Putting x = 0:

0 = 4 sin θ,      sin θ = 0,         θ = 0

The θ interval that corresponds to x = 0 to x = 4 is then θ = 0 to θ = π/2. Hence,

Substitutions in general. A considerable amount of ingenuity is required choosing a substitution that will simplify a given integral. We have covered some of the most useful transformations. However we have by no means exhausted the possibilities. One is free at any time to make any change of variable that he pleases. If the integration can be performed in terms of the new variable, the result can easily be changed back into terms of the original variable.

Important transformations. The following list gives some transformations and their effects.