-
Notifications
You must be signed in to change notification settings - Fork 38
/
Copy path20.html
378 lines (305 loc) · 12.1 KB
/
20.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
<!DOCTYPE html>
<meta charset="utf-8">
<title>Dodges, Pitfalls, and Triumphs | Calculus Made Easy</title>
<link rel="stylesheet" href="screen.css">
<style>
body{counter-reset:h1 20}
</style>
<script type="text/x-mathjax-config">
MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}});
</script>
<script type="text/javascript"
src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.1/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>
<h1><br>Dodges, Pitfalls, and Triumphs</h1>
<p class="a rotatedFloralHeartBullet">
<p><em>Dodges.</em> A great part of the labour of integrating
things consists in licking them into some shape that
can be integrated. The books–and by this is meant
the serious books–on the Integral Calculus are full
of plans and methods and dodges and artifices for
this kind of work. The following are a few of
them.
<p><em>Integration by Parts.</em><a name="intparts"/> This name is given to a
dodge, the formula for which is
\[
\int u\, dx = ux - \int x\, du + C.
\]
It is useful in some cases that you can't tackle
directly, for it shows that if in any case $\int x\, du$ can
be found, then $\int u\, dx$ can also be found. The formula
can be deduced as follows. From <a href="6.html#differprod">here</a>, we have,
\[
d(ux) = u\, dx + x\, du,
\]
which may be written
\[
u(dx) = d(ux) - x\, du,
\]
which by direct integration gives the above expression.
<p>
<p><em>Examples</em>
(1) Find $\int w · \sin w\, dw$.
<p>Write $u = w$, and for $\sin w · dw$ write $dx$. We shall
then have $du = dw$, while $\int \sin w · dw = -\cos w = x$.
<p>Putting these into the formula, we get
\begin{align*}
\int w · \sin w\, dw &= w(-\cos w) - \int -\cos w\, dw \\
&=-w \cos w + \sin w + C.
\end{align*}
<p>(2) Find $\int x \epsilon^x\, dx$.
\begin{align*}
\text{Write}\;
u &= x, & \epsilon^x\, dx&=dv; \\
\text{then }\;
du &= dx, & v &=\epsilon^x,
\end{align*}
and
\[
\int x\epsilon^x\, dx
= x\epsilon^x - \int \epsilon^x\, dx
\quad \text{(by the formula)} \\
= x \epsilon^x - \epsilon^x = \epsilon^x(x-1) + C.
\]
<p>(3) Try $\int \cos^2 \theta\, d\theta$.<a name="moreexamples"/>
\begin{align*}
u &= \cos \theta, &\cos \theta\, d\theta &= dv. \\
\text{Hence }\;
du&= -\sin \theta\, d\theta, & v &=\sin \theta,
\end{align*}
\begin{align*}
\int \cos^2 \theta\, d\theta
&= \cos \theta \sin \theta+ \int \sin^2 \theta\, d\theta \\
&= \frac{2 \cos\theta \sin\theta}{2} +\int(1-\cos^2 \theta)\, d\theta \\
&= \frac{\sin 2\theta}{2} + \int d\theta - \int \cos^2 \theta\, d\theta.
\end{align*}
\begin{align*}
\text{Hence}\;
2 \int \cos^2 \theta\, d\theta
&= \frac{\sin 2\theta}{2} + \theta \\
and
\int \cos^2 \theta\, d\theta
&= \frac{\sin 2\theta}{4} + \frac{\theta}{2} + C.
\end{align*}
<p>(4) Find $\int x^2 \sin x\, dx$.
\begin{align*}
\text{Write }\;
x^2 &= u, & \sin x\, dx &= dv; \\
\text{then }\;
du &= 2x\, dx, & v &= -\cos x,
\end{align*}
\[
\int x^2 \sin x\, dx = -x^2 \cos x + 2 \int x \cos x\, dx.
\]
<p>Now find $\int x \cos x\, dx$, integrating by parts (as in
Example 1 above):
\[
\int x \cos x\, dx = x \sin x + \cos x+C.
\]
<p>Hence
\begin{align*}
\int x^2 \sin x\, dx
&= -x^2 \cos x + 2x \sin x + 2 \cos x + C' \\
&= 2 \left[ x \sin x + \cos x \left(1 - \frac{x^2}{2}\right) \right] +C'.
\end{align*}
<p>(5) Find $\int \sqrt{1-x^2}\, dx$.
\begin{align*}
\text{Write}\;
u &= \sqrt{1-x^2},\quad dx=dv; \\
\text{then }\;
du &= -\frac{x\, dx}{\sqrt{1-x^2}}\quad \text{(see }
\style{
border-bottom-width: 1px;
border-bottom-style: dotted;
}{\href{9.html}{\text{Chap. IX}}})
\end{align*}
and $x=v$; so that
\[
\int \sqrt{1-x^2}\, dx=x \sqrt{1-x^2} + \int \frac{x^2\, dx}{\sqrt{1-x^2}}.
\]
<p>Here we may use a little dodge, for we can write
\[
\int \sqrt{1-x^2}\, dx
= \int \frac{(1-x^2)\, dx}{\sqrt{1-x^2}}
= \int \frac{dx}{\sqrt{1-x^2}} - \int \frac{x^2\, dx}{\sqrt{1-x^2}}.
\]
<p>Adding these two last equations, we get rid of
$\int \dfrac{x^2\, dx}{\sqrt{1-x^2}}$, and we have
\[
2 \int \sqrt{1-x^2}\, dx = x\sqrt{1-x^2} + \int \frac{dx}{\sqrt{1-x^2}}.
\]
<p>Do you remember meeting $\dfrac {dx}{\sqrt{1-x^2}}$? it is got by
differentiating $y=\arcsin x$ (see <a href="15.html#intex3">here</a>); hence its integral
is $\arcsin x$, and so
\[
\int \sqrt{1-x^2}\, dx = \frac{x \sqrt{1-x^2}}{2} + \tfrac{1}{2} \arcsin x +C.
\]
<p>You can try now some exercises by yourself; you
will find some at the end of this chapter.
<p><em>Substitution.</em> This is the same dodge as explained
in <a href="9.html">Chap. IX</a>. Let us illustrate its application
to integration by a few examples.
<p>(1)
$\int \sqrt{3+x}\, dx$.
\begin{align*}
Let
3+x &= u,\quad dx = du; \\
replace
\int u^{\frac{1}{2}}\, du
&= \tfrac{2}{3} u^{\frac{3}{2}} = \tfrac{2}{3}(3+x)^{\frac{3}{2}}.
\end{align*}
<p>(2) $\int \dfrac{dx}{\epsilon^x+\epsilon^{-x}}$.
Let
\[
\epsilon^x = u,\quad \frac{du}{dx} = \epsilon^x,\quad\text{and}\quad
dx = \frac{du}{\epsilon^x}; \\
\]
so that
\[
\int \frac{dx}{\epsilon^x+\epsilon^{-x}}
= \int \frac{du}{\epsilon^x(\epsilon^x+\epsilon^{-x})}
= \int \frac{du}{u\left(u + \dfrac{1}{u}\right)}
= \int \frac{du}{u^2+1}.
\]
<p>$\dfrac{du}{1+u^2}$ is the result of differentiating $\arctan x$.
<p>Hence the integral is $\arctan \epsilon^x$.
<p>(3) $\int \dfrac{dx}{x^2+2x+3} = \int \dfrac{dx}{x^2+2x+1+2} = \int \dfrac{dx}{(x+1)^2+(\sqrt 2)^2}$.
<p> Let $x+1=u,\quad dx=du$; then the integral becomes $\int \dfrac{du}{u^2+(\sqrt2)^2}$; but $\dfrac{du}{u^2+a^2}$ is
the result of differentiating $u=\dfrac{1}{a} \arctan \dfrac{u}{a}$.
<p>Hence one has finally $\dfrac{1}{\sqrt2} \arctan \dfrac{x+1}{\sqrt 2}$ for the value
of the given integral.
<p><em>Formulæ of Reduction</em> are special forms applicable
chiefly to binomial and trigonometrical expressions
that have to be integrated, and have to be reduced
into some form of which the integral is known.
<p><em>Rationalization</em> , and <em>Factorization of Denominator</em>
are dodges applicable in special cases, but they do not
admit of any short or general explanation. Much
practice is needed to become familiar with these preparatory
processes.
<p>The following example shows how the process of
splitting<a name="partfracs"/> into partial fractions, which we learned in
Chap. XIII., <a href="13.html#partfracs2">here</a>, can be made use of in integration.
<p>Take again $\int \dfrac{dx}{x^2+2x+3}$; if we split $\dfrac{1}{x^2+2x+3}$
into partial fractions, this becomes (see <a href="20.html#partfracs3">here</a>):
\[
\dfrac{1}{2\sqrt{-2}} \left[\int \dfrac{dx}{x+1-\sqrt{-2}} - \int \dfrac{dx}{x+1+\sqrt{-2}} \right]
\]
\[
=\dfrac{1}{2\sqrt{-2}} \log_\epsilon \dfrac{x+1-\sqrt{-2}}{x+1+\sqrt{-2}}.
\]
Notice that the same integral can be expressed
sometimes in more than one way (which are equivalent
to one another).
<p><em>Pitfalls.</em> A beginner is liable to overlook certain
points that a practised hand would avoid; such as
the use of factors that are equivalent to either zero or
infinity, and the occurrence of indeterminate quantities
such as $\tfrac{0}{0}$. There is no golden rule that will meet
every possible case. Nothing but practice and intelligent
care will avail. An example of a pitfall which
had to be circumvented arose in <a href="18.html">Chap. XVIII</a>,
when we came to the problem of integrating $x^{-1}\, dx$.
<p><em>Triumphs.</em> By triumphs must be understood the
successes with which the calculus has been applied to
the solution of problems otherwise intractable. Often
in the consideration of physical relations one is able
to build up an expression for the law governing the
interaction of the parts or of the forces that govern
them, such expression being naturally in the form of
a <em>differential equation</em>, that is an equation containing
differential coefficients with or without other algebraic
quantities. And when such a differential equation
has been found, one can get no further until it has
been integrated. Generally it is much easier to state
the appropriate differential equation than to solve it:–the
real trouble begins then only when one wants to
integrate, unless indeed the equation is seen to possess
some standard form of which the integral is known,
and then the triumph is easy. The equation which
results from integrating a differential equation is
called* its “solution”; and it is quite astonishing
how in many cases the solution looks as if it had no
relation to the differential equation of which it is
the integrated form. The solution often seems as
different from the original expression as a butterfly
does from the caterpillar that it was. Who would
have supposed that such an innocent thing as
\[
\dfrac{dy}{dx} = \dfrac{1}{a^2-x^2}
\]
could blossom out into
\[
y = \dfrac{1}{2a} \log_\epsilon \dfrac{a+x}{a-x} + C?
\]
yet the latter is the <em>solution</em> of the former.
<p><note>*This means that the actual result of solving it is called its
“solution.” But many mathematicians would say, with Professor
Forsyth, “every differential equation <em>is considered as solved</em> when
the value of the dependent variable is expressed as a function of
the independent variable by means either of known functions, or of
integrals, whether the integrations in the latter can or cannot be
expressed in terms of functions already known.”</note>
<p>As a last example, let us work out the above together.
<p>By partial fractions,<a name="partfracs3"/>
\begin{align*}
\frac{1}{a^2-x^2} &= \frac{1}{2a(a+x)} + \frac{1}{2a(a-x)}, \\
dy &= \frac {dx}{2a(a+x)}+ \frac{dx}{2a(a-x)}, \\
y &= \frac{1}{2a}
\left( \int \frac{dx}{a+x}
+ \int \frac{dx}{a-x} \right) \\
&= \frac{1}{2a} \left(\log_\epsilon (a+x) - \log_\epsilon (a-x) \right) \\
&= \frac{1}{2a} \log_\epsilon \frac{a+x}{a-x} + C.
\end{align*}
Not a very difficult metamorphosis!
<p>There are whole treatises, such as Boole's <em>Differential
Equations</em> , devoted to the subject of thus finding
the “solutions” for different original forms.
<p>
<hr><h3>Exercises XIX</h3>
<p>
<p>(1) Find $\int \sqrt {a^2 - x^2}\, dx$.
<p>(2) Find $\int x \log_\epsilon x\, dx$.
<p>(3) Find $\int x^a \log_\epsilon x\, dx$.
<p>(4) Find $\int \epsilon^x \cos \epsilon^x\, dx$.
<p>(5) Find $\int \dfrac{1}{x} \cos (\log_\epsilon x)\, dx$.
<p>(6) Find $\int x^2 \epsilon^x\, dx$.
<p>(7) Find $\int \dfrac{(\log_\epsilon x)^a}{x}\, dx$.
<p>(8) Find $\int \dfrac{dx}{x \log_\epsilon x}$.
<p>(9) Find $\int \dfrac{5x+1}{x^2 +x-2}\, dx$.
<p>(10) Find $\int \dfrac{(x^2 -3)\, dx}{x^3 - 7x+6}$.
<p>(11) Find $\int \dfrac{b\, dx}{x^2 -a^2}$.
<p>(12) Find $\int \dfrac{4x\, dx}{x^4 -1}$.
<p>(13) Find $\int \dfrac{dx}{1-x^4}$.
<p>(14) Find $\int \dfrac{dx}{x \sqrt {a-bx^2}}$.
<p><h3 class="answers">Answers</h3>
<p>(1) $\dfrac{x\sqrt{a^2 - x^2}}{2} + \dfrac{a^2}{2} \sin^{-1} \dfrac{x}{a} + C$.
<p>(2) $\dfrac{x^2}{2}(\log_\epsilon x - \tfrac{1}{2}) + C$.
<p>(3) $\dfrac{x^{a+1}}{a + 1} \left(\log_\epsilon x - \dfrac{1}{a + 1}\right) + C$.
<p>(4) $\sin \epsilon^x + C$.
<p>(5) $\sin(\log_\epsilon x) + C$.
<p>(6) $\epsilon^x (x^2 - 2x + 2) + C$.
<p>(7) $\dfrac{1}{a + 1} (\log_\epsilon x)^{a+1} + C$.
<p>(8) $\log_\epsilon(\log_\epsilon x) + C$.
<p>(9) $2\log_\epsilon(x - 1) + 3\log_\epsilon(x + 2) + C$.
<p>(10) $\frac{1}{2} \log_\epsilon(x - 1) + \frac{1}{5} \log_\epsilon(x - 2) + \frac{3}{10} \log_\epsilon(x + 3) + C$.
<p>(11) $\dfrac{b}{2a} \log_\epsilon \dfrac{x - a}{x + a} + C$.
<p>(12) $\log_\epsilon \dfrac{x^2 - 1}{x^2 + 1} + C$.
<p>(13) $\frac{1}{4} \log_\epsilon \dfrac{1 + x}{1 - x} + \frac{1}{2} \arctan x + C$.
<p>(14) $\dfrac{1}{\sqrt{a}} \log_\epsilon \dfrac{\sqrt{a} - \sqrt{a - bx^2}}{x\sqrt{a}}$. (Let $\dfrac{1}{x} = v$; then, in the result,
let $\sqrt{v^2 - \dfrac{b}{a}} = v - u$.)
<br>
<hr>
<a href="21.html">Next →</a><br>
<a href="/">Main Page ↑</a><br>
<script src="j/jquery.js"></script>
<script src="j/modernizr.js"></script>
<script src="j/dih5.js"></script>
<!-- Google tag (gtag.js) -->
<script async src="https://www.googletagmanager.com/gtag/js?id=UA-101178221-1"></script>
<script>
window.dataLayer = window.dataLayer || [];
function gtag(){dataLayer.push(arguments);}
gtag('js', new Date());
gtag('config', 'UA-101178221-1');
</script>