Integration by Parts

BC Calculus | Post AP: Advanced Integration Techniques
I. Integration by Parts / Reduction Formulas
If u = f (x) and v = g (x) and if f / and g / are continuous, then ∫ udv = uv − ∫ vdu.
Example 1: Evaluate
∫ xe
Example 2: Evaluate
∫ x sec
2x
dx.
2
xdx
Example 3: Evaluate ∫ ln xdx
Example 4: Evaluate
∫ tan
Example 5: Evaluate
∫x
2
−1
x dx
e 2 x dx.
Example 6: Evaluate ∫ e x cos xdx.
Example 7: Evaluate ∫ sec 3 xdx.
Example 8: Find a reduction formula for ∫ sin n x dx
Example 9: Use the reduction formula in Example 8 to evaluate ∫ sin 5 x dx
BC Calculus | Post AP: Advanced Integration Techniques
Integration for powers of trigonometric function (Reduction Formulas):
BC Calculus | Post AP: Advanced Integration Techniques
II. Trigonometric Integrals ∫ sin m x cos n x dx
Guidelines:
Example 1: Evaluate ∫ sin 5 x dx without using “by parts.”
Example 2: Evaluate ∫ cos 2 x dx without using “by parts.”
Example 3: Evaluate ∫ cos 3 x sin 4 x dx
BC Calculus | Post AP: Advanced Integration Techniques
III. Trigonometric Integrals
∫ tan
Guidelines:
Example 1: Evaluate
∫ tan
Example 2: Evaluate
∫ tan
3
x sec 5 xdx .
2
x sec 4 xdx .
m
x sec n x dx
BC Calculus | Post AP: Advanced Integration Techniques
IV. Trigonometric Substitutions
Expression in Integrand
a2 − x2
Trigonometric Substitution
x = a sin θ
a2 + x2
x = a tan θ
x2 − a2
x = a sec θ
Example 1: Evaluate
∫x
Example 2: Evaluate
∫
Example 3: Evaluate ∫
dx
2
16 − x 2
dx
4 + x2
x 2 − 9dx
x
BC Calculus | Post AP: Advanced Integration Techniques
V. Integration using Partial Fractions
Guidelines for Partial Fraction Decompositions of f ( x) / g ( x)
1. If the degree of f(x) is not lower than the degree of g(x), use long division to obtain the
proper form.
2. Express g(x) as a product of linear factors ax + b or irreducible quadratic factors
ax 2 + bx + c , and collect repeated factors so that g(x) is a product of different factors of
the form (ax + b) n or (ax 2 + bx + c) n for a nonnegative integer n.
3. Apply the following rules.
Rule a: For each factor (ax + b) n with n ≥ 1 , the partial fraction decomposition contains
the sum of n partial fractions of the form
An
A1
A2
, where each numerator Ak is a real number.
+
+ ⋅⋅⋅ +
2
ax + b (ax + b)
(ax + b) n
Rule b: For each factor (ax 2 + bx + c) n with n ≥ 1 and with ax 2 + bx + c irreducible, the
partial fraction decomposition contains a sum of n partial fractions of the form
An x + Bn
A1 x + B1
A2 x + B
, where each Ak and Bk is a real number.
+
+ ⋅⋅⋅ +
2
2
2
ax + bx + c (ax + bx + c)
(ax 2 + bx + c) n
5x − 3
dx .
− 2x − 3
Example 1: Evaluate
∫x
Example 2: Evaluate
∫ ( x + 2)
Example 3: Evaluate
2x3 − 4x 2 − x − 3
∫ x 2 − 2 x − 3 dx
2
6x + 7
2
dx
BC Calculus | Post AP: Advanced Integration Techniques
VI. Integration using Table of Integration
Example 1: Evaluate
∫x
Example 2: Evaluate
∫x
1
2
3
3 + 5x 2
cos x dx
dx for x > 0
BC Calculus | Post AP: Advanced Integration Techniques
VII. Irreducible Quadratic Expression in the Integrand
Example 1: Evaluate
∫x
Example 2: Evaluate
∫
2
2x − 1
dx
− 6 x + 13
1
x 2 + 8 x + 25
dx
BC Calculus | Post AP: Advanced Integration Techniques
VIII. Miscellaneous Substitutions (Integral contains an expression of the
Example 1: Evaluate
∫
Example 2: Evaluate
∫
x3
3
x2 + 4
1
x +3 x
dx
dx
n
f (x ) )
BC Calculus | Post AP: Advanced Integration Techniques
IX. Miscellaneous Substitution (Integrand is a rational expression in sin x and cos x )
Example 1: Evaluate ∫
Example 2: Evaluate
1
dx
4 sin x − 3 cos x
cos x
∫ 1 + sin
2
x
dx
482 llll
7
CHAPTER
TECHNTQUES OF TNTEGRATTON
13.
fax
,--.
Jl x--bx
15.
l-dx
J3
x' -
1^,
14x'-lx'-4
,,.
,r.
J
y(y+2)(y-3)
_
1
(x+5)'z(x-1)
)-.
)dx
x3+4
T -I L
x'*3x12
^uL
'r.
lJ
,0.
I (x+a)(x+b)
_
,t.
,0.
_
)-
1
or.J _.1
)*
-Y'
-
or.
ro.
-T
-dx
x'- 5x + lb
j (2x+7)(x-2)2
dx
j -dx
x2-x*6
x'*3x
,o.l
Ej
,,. 7 x'+x+1,
(x' -t 1)'
-dx
,,.
x--ZX-l
),t. J _(x-l)'G'+1)
sr.
ffi sl.
7^
T
,3x2+x+4
l-dx-i3x-t./.
ft
1
l-dx
-Jx'-1
rt x3+2x
33. l---dx
Jo.r*+4x'*3
35.
rdx
.=)I x\x'l-
)^
r x'--1x+
I
J =---dx
(x'-4x*6)'
39-50 Make
116
41. I ^E dx
Jg x-4
x3
WJ
-r rI -;-dx
Jx.+ I
r1---=------;-r
rI \/x-lx dx
r
*
^/l
l-d*
)-.
qi
sec2/
tan2t+3tal t+2
J
(e'-2)G2'+t)
e'
J
m(x'?
-
dt
dx
x + 2) dx
i?. I.tan-txdx
Us" a graph of f(x):ll|'
- 2x - 3) ro decide whether
lif G) a* is positive or negative. Use the graph to give a rough
estimate of the value of the integral and then use partial fractions
to find the exact value.
, x4+3x2+l
l-dx
x'* 5x'* 5x
)*
I
J
2x2) and an antiderivative on the
r"j
dx
----_
x'-2.x
/r\
cos[;):---
the integrand as a rational
VI
: $i.]
t
and
L
1
1L
-,2
cosx:-:------,
\+t'
.2t
stn.r:
and
(c) Show that
1
[rrinr; Substitute u
/r\
,tr\r/:ffi
1
(b) Show that
rdx
40. lJ2Jx*3ix
rr
47. | . _i_
.1-dx
JU I , VX
"" !^r: ax
qq.l'
Jtlz x' * x
2x-l
+ 12x -'7
1
4x2
57. The German mathematician Karl Weierstrass (1815-1897)
noticed that the sribstitution t : tan(x/2) will convert any
rational function of sin x and cos -r into an ordinary.rational
function of r.
(a) Ifr: tan(xlz), -r I x ( .r, sketch a right triangle oruse
trigonometric identities to show that
\L/
a substitution to
@jqo.
rx
-t
J
rx3+2x2*3x-2
I _
J (x'*2x*2)'
"xp{ess
function and thbn evaluate the integral.
45.
155.1
I -J- a,
J-r'*1
17, I
x
J
x
l-dx
x'44x*13
36.
sln
55-56 Evaluate the integral by completing the square and using
Formula 6.
Jo
4)'
-I
same screen.
J x
l3l.l
-r
t
ffi Sl.- Craph both y : 1lG' -
-AX
x*4
l79.ll"--dx
-r x'*2x15
-J
)-
5 l-52 Use integration by parts, together with the techniques of this
section, to evaluate the integral.
ds
,,. T --r-------ls-l,s
1l'
y',
COS
sm:x
x'-x-6
x2+2x-1
_)-
J
e2'
e"*3e'*2
x3-4x-70
l6'Jn
2x'
4v2-l.v-12
"
Llx l,
'r.
dx
x-1
_
l
^1
lll.
l--dx
*Jzx'-7
-
)uL-
2
1+t'
-
)+
1+t' ^uL
Use the substitution in Exercise 57 to transform the inte'
grand into a rational function of f and then evaluate the integral.
58-61
58.
rdx
l-
J 3-5sinx
./x
3sinx-4cosx
60.
l't'
J.B
I
1
+ sinx-cos,
d]
Multivariable Calculus O9l0: Unitl
-
Principles of Integrations (Unit Exercises)
Principles of lntegral Evaluation
A convergent improper integral over
an
Approximate this integral by applying Simpsou,s
rule
with 2n : 20 subdivisions to the integral
infinite interval can
be approximared by first replaciag the infinite limit(s)
of in_
tegration by finite iimit(s), then using a numerical integration
technique, such as Simpson's rule, to approximate the integral
f4 1
I
Jo xb+7
Round -dx
your answer to three
with finite limi(s). This technique is illusffated in Exercises
60 and 61.
60.
decimal places and compare it to n/3 rounded to three decimal places.
Suppose that the integral in Exercise 5g is approximated
by
first writing it
G)
as
rK
f+*
Use the result that you obtained in Exercise 46 aad
the
fact that ll@6 + | < l/x6 for x > 4 to show that
the
truncation error for the approximation in part (a) sads_
r*a
| ,-"" dx = Jo
| "-r' d* + Jx
I e-,' dx
Jo
fies0<E<Zxl0-4.
then dropping the second term, and then applying Simpson,s
rule to the integrai
tK
L-"
d,
Jo
The resulting approximation has two sources of error: the
error from Simpson's rule and the error
r*a
E: I
JK
ffi
62. For what values ofp does [*lo
"rdx
63. Show ,au,
< 1 anddivergesifp > I.
64. It is sometimes possible
that resuits from discarding the second term. We
truncatian etor.
2n:
an
Uy
making the indicated substitution, and investigate whathap_
if you evaluate the integral directiy using a CAS.
cal E the
pens
(a) Approximate the integral in Exercise 5g by applying
raI
to convert an improper integral iato
a "proper" integral having the same value by making
appropriate substitution. Evaluate the following integrat
e-,'dx
Simpson's rule with
['4xp "oru".g"rifp
Jo
converoe?
al
lg subdivisions to the inte_
I *-x
:-
J,
b^q
dx;
u=$-1
t3
L-"
d*
Jo
_
Round your answer to four decimal places and compare
j.,8 roundeO to four decimal ilu."..
Use the result that you obtained in Exercise 46 and the
it to
.
(b)
factthate-xz
S lxe-* forx > 3 to show that the fun_
cation error for the approximation in part (a) satisfids
0<E<2.1x10-5.
61. (a) It
65.
can be shown rhat
[+IJo r0+1
-d*:!
1
66.
3
1. Consider the following methods for evaluating
integrals:
a-substitution, integration by parts, partial fractions, reduc_
fion formulas, and trigonomefic substirutions. In each part,
state the approach that you would ry first to evaluate
the
integral. If none of them seems appropriate, then
say so.
You need not evaluate the integral.
O,
|
,sinxdx
f_
(c) ltan'xdx
| ^)
-7X'
(b)
I
le) I
J x'*7
-dx
sinxdx
/"ou
fal
It^'xsec2xdx
rrt
1ff6a.
/,1 cos
x
J, "n dx; u: Jx
ft sinx
I --_dx;u-Jl-x
Jo t/ I - x
{e)
|t_
an-'*a*
(D
(l ,Jq - x2 dx
.l
r_
JJt-xzdx
2. Consider the following trigonometric
substitutions:
x:3sin6, x-JtznQ, x:3sec0
In each part, state the substitution that you would fy first to
evaluate the integral. If none seems appropriate, then state
a trigonometric substitution that you would use. you need
not evaiuate the integrat.
f_
@I
r
,/g*x2dx
"J"l6-dx
@)
! ,E-u'0,
at I Jr*u'0.
at
[ ,t.'- ro.
(d)
14. Find apositive value ofa
f-
+ (ex)z dx
J Jt
(f)
3. (a) What condition must a rational function satisfy for the
method of partial fractions to be applicable directly?
@) If the condition ia part (a) is not satisfied, what must
you do if you waat to use partial fractions?
4. What is
I
of the formula ia the Endpaper
you
apply to evaluate the integral.
would
that
Table
Integral
You need not evaluate *re integral.
|
(c)
l
(e)
dr
.J' - .'o'
*
;t
-1i,:
ii
,fi
A'
fdx
,_
I
I
-
xJ4x
7. In each part" evaluate the integral by making
I
*, I
snauax
fl
the integrat I -;- J xt_x
an appropriate
Evaluate the integral using the substitution
For what values of .r is your rcsult valid?
x:
sec 6 '
:
siaO'
x
Id,
I xiz*x*1)
zG.
-
J+
['Go*
x*9
J6
I J*-ta*
^Ia2
zB.
Jo
x dx
dx
;,-TW,
a.b >
o
!
[-Lo,
I Jx_xz
three ways: using the substitution u : G, using the
substitu[on a - JT1, and completing the square.
(b) Show that the answers in part (a) are equivalent.
Find the area of the region that is enclosed by the curves
) = (x - 3)l (x3 + xz), y -- 0, x : 1, andx : 2-
sketch to show that
I#,'
dv
I Some htegrals that can be evaluated by hand cannot be eval- i
! uated by all computer algebra systems. fu Exercises 31-34,
I evaluate the iategral by han4 and determine if it can be eval- i
I uated on your CAS.
.. - ---- ..-- L .-.-...
I
m31.
l*- , .-4"
1t -2x213/2 dx
i
Evaluate the iategral
11. sketch the region whose area r,
Ji
)a
Jxz +2x +2
'o' /.
i--"-'-
(c) Evaluate the iategral using the method of partial fractions. For what values of x is your result valid?
(a)
-tt
@T*
dx.
(b)
I
I
I
1'+-
Evaluate the integral using t}le substitution
For what values of x is your resuit valid?
r1:
J sia'0-6sa0*12
f x-13
r+,,.J,
,coss1*2;dx
(a)
'i):
cos0
---
['F x o,
27. [ -La.
J Je'*l
substitution and applying a reduction formula'
fal
J
I @-1)(x*2)(x-3)
JVfi.
/
18.
sec'7x'} dx
fdx
zs.
'.,i
l'.r
2t.
/1
Consider
r.'
tf
*3
t^r' e ae
!0"/o
fdx
,_
J {3 + *ryrt,
xsles' dx
,l'
.ii.
19.
I'#0.
(f)
(a) integration by parts
(b) the substitution u:
i.:ii.l
I
(d)
| "o'a*
6. Evaluate tne int.fa
#
Al
swxcos 9x
equation
rrlJl
f^
t7. I xtanz(xz)
an improper integral?
rc.
/cos 0 s:Jr-? d0
J
5. In each part" find the number
@
Lhat satisfies the
rt' i
I x/*az
Jo
15-30, evaluate the iategral.
-dx:l
and use
your
d' : [' aio,
[*
lo l+x'z - Jt
1#,.
Eil 32. / t"or" x sioro r
-
cos30,
stn32
x) dx
,:----:
B 33. J,l. - J*z - +ax.lfiint: +(Jx +2- 4E -Dz :?l
m 34' I #*dx.fHint:Rewritethe
x10(1
ffi
denomi:rator as
+ x-e).1
35. Let
f(x):
-zx5 +26x4
x6
-xs -
*
18xa
15x3
+ 6xz +aox +
43
-2r3 -39x2 -x -20
(a) Use a CAS to factor the denominator, and then write
V
U. Find the area that is enclosed between the x-axis
curve y : onx
- l) / xz for x Z e.
and the
13. Find the volume of the solid that is generated wheu the region between lhe x-axis aad the curve y
e-' forx > 0 is
revolved about the y-axis.
:
dowu the form of the partial fraction decomposition.
You need not filrd the values ofthe constants.
(b) Check your answer in part (a) by using the CAS to find
the partial fraction decomposition of /.
(c) Integrate / by hand and fhen check your answer by
integratirg with the CAS.
36. TheGarnmafunctian,
f (x)
f (x),
1+o
: 1 (-te-l
is defined
where 6
as
dt
Jo
It
can be shown that this improper iategral converges
onlyifx > 0.
(a) Find f (1).
(b) Prove: f (x * 1)
if
and
: xl(x) for aII .x > 0. [;1inr: Use
iategration by parts.l
(c)
(d)
Use the results ir parts (a) and (b) to find f (2), f (3),
and f (4); and then make a conjecture about I-(n) for
positive integer values ofn.
Show that f (+)
,/i.fHint: See Exercise 58 of Sec-
:
tion
(e)
5t.
Refer to the Gamma function defined
rl
dx
i-u
evaluated by nrrmerical methods.
(a) Obtain (2) from (1) by substituting
cosd
xn . Use the
Exercise 36 to show
:
sin(00/2)
and then making the change of variable
n > 0.
sin
o.
: I -zsnz@/z)
cosdq: I -Zs1n2@o/2)
O)
n>
Q)
where k
sn(00/2). The integral in (2) is called a complete elliptic integral of the fvst kind arrd is more easily
k
= (-l)"1(n * 1),
fHint:l*tr: -Inx.]
+ 1),
(b)
[** ,-,' a, : p(' n
" Jo
\
)'
Vlint: Let t --
d
:
sin(0
/2)lsn(0012)
:
slr:(O
/2)/k
Use (2) and the numerical integration capabiliry of your
CAS to estimate the period of a simple penduium for
which
I:
1.5
fr
0o
:
20o,and g
-
32ftl
s2.
result i:r Exercise 36(b).1
s8' Asimple pendulznz consists of a mass that swings in a vertical plane at the end of a massless rod of length .L, as shown
in the accomFanying figure. Suppose that a simple pendulum is displaced through an angle 06 and released from rest.
It can be shown that in the absence of friction, the time I
required for the pendulum to make one complete back-andforth swiag, calTedthe period, is given by
'r-
[t,l.rt" :)a
l-l
Y s Jo
/cosE
I
-
cos A;
--
(1)
Railroad Design
our company has a contract to construct a track bedfor a railroad lirue between towns A and B shown on the contour
map in Figure L The bed can be created by cutting trenches through the surface or by usirLg some combination of
trenches and tunnels. As chief engitleer, your assignrnent is to arlalyze the costs of trenches an"d. tunnels and to propose
a design strategy for minirnizing the total construction cost.
sw
Engineering Requirements
The Transportation Board submits the following engineering requirements to your company:
.
the
1
J, mrado
':oli
:
: i"fi *dr(i): ifi.
(a) / (t *)'
Jo
ffi
frvn
Use the results obtained in parts (b) and (d) to show that
that
0(r) is the angle the pendulum makes with
can be shown that the period can be expressed as
8.8.1
f (i)
:
vertical at time l. The improper iategral in (1) is difficult
to evaluate numerically. By a substitution ouflined below it
The track bed is to be straight and 10 m wide. The grade is to increase at a cotrstant rate
from the existing elevation of 100 m at town A to an elevation of 110 m at point M and then
decrease at a constant rate to the existing elevation of 88 m at town B.
A7,ffi
The formulas below are stated in terms of constants a, b, c, m, n, and so on. These con_
stants can usually assume any real value and need not be integers. Occasional limitations
on their values are stated with the formulas. Formula 5 requires n * -1, for example, and
Formula 11 requires n * *2. The formulas also assume that the constants do not take on
values that require dividing by zero or taking even roots of negative numbers. For exampie. Formula 8 assumes a * O, and Formula 13(a) cannot be used unless b is negative.
,.l"dv:uv-lrn,
z.
lna
I o. a,: !*c,
,./.o, udu:sina*c
a.
I
*,
'. I
,.
|
r.
[
+ b).
dx:'Tii'ii' . r, n * -l
*r," + b)^ dx -
@x
+-!)*11+*
*t"* + b)-t dx : L -
9')xhx+dzdx=
4rrbx
ll,n
o,L
-
sin, du
a*1, a>o
:-cosz*C
e.[(ax+u) -tdx-_Llnlax+bl-C
a
1
#).
,, n* -7, -2
+ bl + c
d* ,:1rn l--z l * .
n. I x(ax+b)
b lax+bl
J
I
,,, a v11--!
or*)*c
f
21{or46y+z * C, n*
11. I (\'ax -f b)" dx:----*
-2
-a
n-t2
J'
-
f dx
r l\,6-+b-{Ot
|
' )xYax-tb
\/b lYax-rb+Ybl
-:---lnt {,r+n
dx
ol
,0. l{r--)odx:
-iJrt6o-c
J x'
-t+C.
151
n. luG-tb
x d*:2rGr+a
J
if
+
ul--!J x\/ax+b
b>0
_a l---4!---L^
,r. l___!_=_{4*u
bx -2b)*{o*+b
Jx2\/ax+b
rc. I a'lx'
) ^d*
,r.
zo.
^:lt*-,
a
L+ c
a
t7.
l--lt-:l,n1x+ql+c
./.a lx-al
Ja'-x'
l-L:
J\/o2+*2
sinh-l
L+
a
c:
tg.
In (x
+
lE + *') + c
l--!1-- -
J \a' t x')-
[ ' ^d*x')'
,,^ :
) la'
-
xlx
Zaz(az
r
{^n
a *21 ' 2a3 'o
xlfa*
2a2(a2
r_+c'
a
I-
- x2)
2az J az
-
x2
Section
zr.
zz.
f---2
)f
xl
a, +
t-
rt7
)
+ x, a*
\/E+?
f
23.J
_
26.
dx:;\/Cr+7 + ir"1* + tG\
:
+
!la2
z*'){7 + *, -
^4
f
xzt + c
r,
1*
+
t/il
l--L:-1
J x\,/az+xz
I
sin-r
L+
a
c
29.
lx \F-p
* I #:
u.
lz.
cosh-r
f-_^2
t/-, - r, a.
J
:
X* r:
h l,
+!*
)
f*17
J
r
)
-
3s.
(f *z
dx: \jL -
"21'-z +
J
l=-L)
- o,l+ c
orl + c
xz- az)'-'zax.
n*
c. n * -2
oz1{*z - oz -}m l* + {*'
r.r
dx:lx2-a2-asec I ;l*,
t f^z .: , oz ,--7 d*:.x,va'-*-*
2"n'L+C
x2\/ az
|t/-, - r, - Tr"l* + f*, -
- r' ax: {ox2 -
v?-,
.
o')'
I t1r,
-.
f-z
a2 tx
I ^r.
3. | __.:_
ax: Isint:-1xYa2-x2+C
J \./a2 - *2
I(\/F-,,)'a*:N#L #l(f
x\x'z
-c
)1 I
I
L
+?7
r,
d* --ffi
a'x
J x2\,/ a2 + xz
a ,nlo*@l*,
x
I
Jx'ax
r
r ta*f;'1_ x'zl
'
34. l-!!-:-rml"___]_:_____^l+c
x
a I
J x\,/a2-x2
40.
lntegrals
I
I t7:7
38.
A Brief Table of
t o+{oz+*2
dx:\/aztxz-atnj
,
l+c
x. J[-L:
\/a2-x2
tt.
A7
^4
- *l
+c
-l
^/v a'xz
1
a'a
x'
+C
613
_ sin-,
*. J| --:4-:
1/2o*-*z
It
56. I sin ax dx
a sin-r
: -1.or
1.u-)
\ a I-
*
l=:).,
yTax
-
la*l
c
x2 +
C
ur.
f
x sin 2ax
ss.Jsin2axdx:i--;-,
sl.
Ja
ax
oo. Irin,
] f .rr,- 2axdx
axdx:- sin'-laxcosax
+nna
n J-
ur. [.or,
)
ax
J
I
f
{b)J
I
dx:
cosn-l ax sin ax
nu
*
cos(a
sin
+
n
- 1|
n J
b)x
sin (a
b)x
a
dx: L
sit ax
/
cos(.a
-
b)x
stn (a
-t b)x
+ ,1o-y,
+
c,
a2+
"o,
/.or'
cos'-z ax dx
sllr.,a
b\x sin (a * b)x
axsinbxdx:-r; _- U --1G +U
(c)JcosaxcosbxO*: ,**b)
J x\/Zax - x2
b2
+C, az+b2
ax
ax dx
:
+c
t *'#
-,
A7
Section
63.
65.
66.
68.
69.
70.
7t.
n1
74.
75.
76.
: -
,in n, cos d.r a*
/
99!2!*
*
lntegrals
sin'-lax
f
-,
j sinn axcosaxiv: (n* L)a - 5.
64. I
6
n
515
* -l
l+u- d*: I rn,sin axl -r c
J SI1AX
A
I
ax t
C,
I cos'ax sinaxdx: - "osn-t
i*"',--"
(n-t t)a
J
I
I
sinn
ax
cosm
ax dx
:
sinn -t ax
-
n
* -l
cos'-t ax
n
67.
- | L
tin o*
IJ COSdxdx: -! a ln l.o, axt + C
.^ "
Jffi-;i)sinn_2axcoS*Qxdx,n*-m(Ifn_-m'useNo.86.)
f ,in,o, cos*axo*_
sin'-tqxc,os^-tax
atm+n)
J
-+fm-rnJ
,in,rrcos,-2 axdx, m*-n (ffm:_,n,useNo.g7..1
"+-,,,-, [8,,"(+ -+)]-r,
Irr*^:
a
l-t'
,' Ir:k;:L,un(+. +)- ,
Ii*:-:,*(; -T).,
or : --Z:,un-, I E* gl * ..
IJ b+ ccosdir
2J
a\/b2^c2
LYa+c
I
I
a*
)
t
: \sin
ax
-L
"o,
,. * :l
f
,' ,t,, ax dx : - t
82.
[
84.
Irun'o*a*:!tanax*xtC
|
J
88.
/
90.
/
:!
,un' o* a*
,""
,".'
,,
a*
ax
tn lsec axl
ln lsec ax
dx: L
tan ax
D. I *
+c
cos ax
t
+c
b1
xn-.
il.
cos ax ctx
ss.
ss.
2
ax
cos ax
)a-a
+c
: 'T'-t o,! - f ,un'
a(n-I) J
:!
f
<
c2
'2
80.
a*
cz
o* : -l ,or9! + c
,r. I l-cosax
a
J
*
,un o*
>
|
o- : l,un !'
IJ l*cosax
o Z+C
rino* a*
bz
lc+bcosax+\F-b2sinaxl + c'
-;a,*: hr'1ff
)a'a
b2,c2
lr+bsino*+\7r2-b1 coso*l ^ o-<-c'
,, b*csinax
I a*
-t
)n-rt't"*:;ffiInl
78. I
86.
A Brief Table of
dx.
n
*
I
*"
|
J
dx: 1.o,
cos ax
.ot ay dx
J "ot'
sz. [.ot,
J
ax
+c
89. I
csc
Ja
lr.
/"r"'
: t
:1tn
dx:
-!
sin
Lsin ax
+
-
c
xn-t sin ax dx
". - :[
lrin axl +
*
"orax
c
x
tC
axdx:_ co:r'-ta.!
aln_t) - Jf .or,,-, axdx, n*1
lt
tan axl
dx
ax
axdx
: -a1n lcscax -
axdx: _.L"otax*
C
cotaxl
*C
92.
%.
94.
I
sec'ax dx
I
csc' ax dx
I
_
secn-z ax tan
a(n
l)
-
ax L
n-21I
N*IJ
sec"-z ax
: -- cscn-z ax cot ax * n-21
a(,r - I) n-lJ- I
- sec'ax
sec'axtanaxdx:-+C
csc''-z ax
n*l
dx, n*1
,
n*0
C,
na
dt ,
es.
tz.
r
98.
J
tan-1
axdx: xtan-lr. -
ss. | *^sin lax
J
,*.
*h
(1 + azxz)
d*: J::sin-rax -;
axdx:
+
rc2.
704.
cos-r
I
xe'-
106. i x,bo,
J
fir.
fe"
.n,
4x::,
(ax
#cos-,cx .
-
l,o,
l
axdx:ircos-t
-
n
I
115.
r17.
-
* r, b> 0,
b
b>o' b+l
'rr=b)xn-tbaxdx)
sin bx dx
:
#
(a sinbx-
b cos bx) + C
eo* cos bx dx :
(a cos bx* D sin bx) + C
109. I lnax dx :
;#
/
J
xn-t(ln ax)^ m f
I
110. x'(ln ax)* dx: --#ixnfin ax)*-t dx, n* -1
-;;
J
I
113.
- !\/l
a
f
lln ax\'-l
x-r(lnax)*dx:-; n, + C. m*
J
ltsl*axdx:
J
ry
- x*
C
tn ltn axl
+c
Icosh ax a*: lsinh ax * C
Ja
f cosh2 ax dx : sinh 2ax x
116.
-,
ff
J
fi4.
-coshax*C
J
:
xlnax
f,
w. | -:!-:
J X tlrAX
-l
t sinh'zaxdx:,2;a-i*,
. "
sinh2z
/ ,mn, ax dx
-
+ I
a2x2
t -l
,*.
111.
I-
ax
1
nI
ro5. I x,eo* dx: *,
a
J
"--;)xn-leaxdx
C
,
dx: *nbo*
alnb-
n+0
rt*,-1
lo3. I bo'dx: lT
a lnb
J
l)
ax
nr -l
r[ #,
leo*dx:leo,-C
Ja(
f
cscn
C
r=l i#,
ror.
axdx:#*"-, o,-fi1 #
/x,ran-,
/x,
-
na
/",. ''axcolaxdx:---+C,
sithn-z ax dx,
n*0
+
C
+l
-
C
Section
A7 A Brief
Table of lntegrals
+n- |
coshn-t axsirthax
lcoshn-2axdx, n*0
I
tt9. xsinhax d*:lcoshax -{sint ax* C
I
Ixcoshaxd*:Lsinhax-{"ort
axdx:4rirt
coshaxdx
tzL. x'sinh axdx:4.orh o*-Ll
a or-Ll *'
a
aJ
)
I
f
a, : lln
axl + c
+c
a* : Ltn
I
aJa
I
: x- 1,*
*
*C
dx: x- 1
I
a-Ja
I
: - t?nh'-! .ax + |
dx, *
I
ln-l)a
'
axdx, *
axdx: - -..--',.o* + |
I
ln-l)a
)
9'
:f
*C
a I Tl*,
J[ "r.r,
I a* : oLsin ,frax)*C
I
-l C
*
a* : -L
d* :
I
a-Ja
t
ax
+ " - I s"rh'-2 d*, *
o* (n-l)a
n-l)
I
csch'-2axcothax n-21
n*
t34. cschraxdx: _
--------=
n-LJ | csch,'-'axdx,
\n-l)a
I
136. Icschnaxcothaxdr:-"t"ho* +
dx:-sechnax *r, n*o
nqJna
t
+
'-0. 1* ,.
z 1al_b- a-b)'-'
I sinh bxdx: +l-L
.o*I
+
e*cosh bxdx:'l_^ |'o' . +'-",1 + C,
2la+b a-b)
I
118.
coshnaxo*-
nanJ
+C
120.
122. lxncosh
xn-t
(cosh ax)
123.
tanh ax
L2s.
tanhz
127.
tanhn ax
ax dx
ax
tanh,-z ax
n
1
cothn-z
n
I
dx
124.
coth ax
126.
cothz ax
aJ
'S inh ax dx
lsinh
coth ax
C
J
cothn
L29.
sech ax
131.
1.33.
sechz
c
coth'-
128.
I (tanh ax)
130.
2,
sech'.-z ax-tanh
sechtr ax
ax dx
.jJ2.
tanh ax
ax
o*
n
m l,unn
I
coth ax
cschz ax
C
I
I
sech'axtanhax
135.
t37.
eo,
az
b2
az
b2
I
138.
*
t39.
I *,-rr-, dx:
(n
- l)1, n) 0
,*.
t 1.3.5 "' (n- l)'ff
t{t.
lo''' "n.
* o*
:
ln'''
cosn
x dx
:
2.4.6'..n
I
617
Z.
4.6 ... ln _
3.5.7 ..-n
c-n,' dx
ifnisaneven
2'
1)
f
irteger >- 2
if n is an odd
integer > 3
:
+rE
a)
O
C,
n*0