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Unformatted text preview: N.1 Ilnteérationj _T_h_e substitution gm Apostol proves only one version of the substitution rule, the one given in Theorem 1
following. Sometimes the converse is needed; we prove this result in Theorem 2. Theorem L Assume Ea; f(u) gig g(x) and g’(x) a_r_e continuous and .t_l_1_a._t f(g(x)) i_s deﬁned Q _a._ll x in m domain of 3. Li Jf(u)du = P(u) + C, m Jr(g(x))g'<x)dx = Peon» + a.
Proof. We are given that P'(u) = {(u). The chain rule implies that the deriVative of P(s(X)) equals P'(g(x))g'(x) = r(g(x)_)g'(x).
This is just the desired result. 0
flfhegrem _2_._ M converse! mum ﬂit f(u) and g(x) and g’(x) are Mm,
mnmmmmmmnxnmmm mammal: = g(x) 1mm differentiable inverse function x = h(u). I_f
W 4 .. j f(g(x))g'(x)dx = QM + C, he;
so . j r(u)du = Q um) + C . Proof. Applying Theorem 1, we can substitute x = h(u) in the given formula (*) to
obtain the equation I &(g(h(u)))g’(h(u)]h’(u)du =. G (Am) + C . Because h is the inverse function to g, we know that g(h(u)) = u and M“) = 1/s’(h(u)). This formula thus takes the form Jf(u)du = Qua.“ + G. which is the equation we wished to prove. a N.2
Example _1_. The usual application of the substitution rule uses Theorem 1. One
begins with the given integrand, and tries to write it in the form f(g(x))g'(x) for some
suitable function f and g, where f is a function we know how to integrate. For example, suppose we wish to compute the integral J x2 cos(x3)dx. 2 2 We see this is almost of the form J cos n du if we set 11 = x . That is, we "group x with dx and supply a factor of 3", writing the integral in the form
3,1, I cos (x3)[3x2dx].
Then because we know that J cos u du = sin u + C, we conclude from Theorem 1 that our
given integral equals
ésin(x3) {L C.
Example 2. On the other hand, sometimes Theorem 2 is the one that is useful. It
often applies when there is nothing obvious to "group with the dx" to simplify the integrand. Trigonometric substitutionsare of this type. For example, consider the integral] f(u)du, where f(u) = }/'u/1l_u2 . This is not something we know how to integrate. However, the substitution u = tan x twill simplify" the expression J 1+u2 at least. It is an acceptable substitution, since it has the
differentiable inverse function x = arctan u Using this substitution, we have 2 2 2 1+u =1+tan x=sec it. Then 1+u =sec’x; N.3
the sign is +, because x lies between —w/2 and + 1r/2, so sec 1: = 1/cos x is positive.
And of course we have
7’ 2
'4'“, = sec x.
Hence the integral j (l/WMu takes the form
J (1 /sec x)sec2 x dx
which we know how to integrate. Indeed; Isecxdx=logsecx+tanx +C. Then we can apply Theorem 2 to conclude that I [1 J 1+u?) du = loglsec(arctan u) '+ tan(arctan u) + C. This answer can be written more simply. For if x = arctan u, then sec 1: = 1+u , as noted earlier, and tan x = 11. Hence we have the formula. J[}/m]du=loglm+ u + C. N.4 A strategy for integration. Step 1. Determine whether you can simplify the integrand easily, using algebraic manipulations (such as completing the square), trig identities, or a simplifying substitution (especially for the "inside function" in f(g(x)).) Examine the form of the integrand to determine the appropriate method. (a) A product of two dissimilar functions suggests integration by parts. [Examplesz x2 sin x, xex.] The same holds for a function whose derivative is a more familiar func—
tion than the function itself. [Examples: log x, arctan x.]
(b) Rational functions of x can always be integrated by the method of partial fractions. (c) Powers of trig functions can be integrated using the half—angle formulas, various substitutions or (if necessary)
reduction formulas. (d) For integrands involving YaZxz, 'Yg§:;3: ‘sz—az,
a trig substitution is often helpful. I (e) [Optional: Any rational function of sin x and
cos x can be reduced to a rational function of u by means of the substitution u = tan(x/2).] Exercises Evaluate the following: 1. I cos2 x dx. (Use either a halfangle formula derived from the identity cos 2x = 1 — 2 sin2.x 2.c032 X  1 I II or the reduction formula of p. 221_) 1/2
2. f /1  “2 du. 0
du . l .
3. Integrate —————7 by expreSSLng ——~——7 in the
l  u 1 ‘ H
form + b Use this formula to evaluate
lu 1+u' J sec x dx = I 99§_§_§§ = I . cos x dx
___—___7T‘“ . cos x l  sin x
4. I d“
112  1
5. I1 d”
3 .
0 (/u2 + 1)
4’ Compute . 1
J xf"(2x)dx,
0
given that f " is continuous for all x, and
{(0) = 1 , f’(0) = 3.
{(1) = 5, f’(1)= 2.
{(2) == 7 , f’(2) == 4.
x4 + 2 I '
1' Evaluate I —Z———3————2 dx completely, including the constants
x + x + x A, B, in the partial fraction decomposition. ...
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 Spring '09
 BRUBAKER

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