MTH 252 – (Integral) Calculus
Chapter 8 Review
1.
Integration by recognition. See list of formulas on pages 511-512 (you do NOT need to know
formulas 23, 27, and 28). Add to this list:
 tan x dx  ln sec x  c   ln cos x  c
 sec x dx  ln sec x  tan x  c
 cot x dx   ln csc x  c  ln sin x  c
 csc x dx  ln csc x  cot x  c




2.


u
c
a
a u
du
u
#25 - 
 cosh 1  c
a
u 2  a2
du
1
u
 tanh 1  c
#26 -  2
2
a u
a
a
#24 -

du
2
2
 sinh 1
Integration by substitution.
 f ( g ( x)) g '( x)dx
=  f (u ) du
3.

Let u  g ( x)  du  g '( x)dx
Integrate and then substitute back to get the answer in terms of x.
Integration by Parts.
 udv = uv   vdu
i.e.

Let u  f ( x)
f ( x) g ( x)dx
&
du  f '( x)dx &
dv  g ( x )dx
v  G ( x )   g '( x )dx
= f ( x)G ( x)   G ( x) f '( x)dx
4.
Trigonometric Integrals (i.e. Powers of the trigonometric functions)
 sin
m

If m is odd, put a sinx with the dx, change the remaining sines to cosines ( sin 2 x  1  cos 2 x ),
and let u = cosx & du = -sinx.dx
If n is odd, put a cosx with the dx, change the remaining cosines to sines ( cos 2 x  1  sin 2 x ), and
let u = sinx & du = cosx dx
If both are even, use the half angle formulas:
cos 2 x  12 (1  cos 2 x) & sin 2 x  12 (1  cos 2 x)


 tan



x cos n x dx
m
x secn x dx
If m is odd, put a tanx secx with the dx, change the remaining tangents to secants
( tan 2 x  sec 2 x  1 ), and let u = secx & du = secx tanx.dx
If n is even, put a sec2x with the dx, change the remaining secants to tangents
( sec 2 x  1  tan 2 x ), and let u = tanx & du = sec2x dx
Otherwise, change the tangents to secants ( tan 2 x  sec 2 x  1 ) and use integration by parts with
dv = sec2x dx (if there are more than 2 secants).
5.
Trigonometric Substitutions
If the integral involves …
a2  x2
x2  a2
Make substitutions using …
x  a sin     sin 1 ax
Or use these substitutions …
x  a sech     sech 1 ax
dx  a cos  d
dx  a sech  tanh  d
a 2  x 2  a 2 cos 2 
x  a sec     sec 1 ax
a 2  x 2  a 2 tanh 2 
x  a cosh     cosh 1 ax
dx  a sec  tan  d
dx  a sinh  d
x  a  a tan 
x  a tan     tan 1 ax
x 2  a 2  a 2 sinh 2 
x  a sinh     sinh 1 ax
dx  a sec 2  d
dx  a cosh  d
x 2  a 2  a 2  x 2  a 2 sec 2 
x 2  a 2  a 2  x 2  a 2 cosh 2 
2
a 2  x 2 or x 2  a 2
6.
2
2
2
Integrating Rational Functions by Partial Fractions:
If the degree of the numerator is larger than the degree of the denominator, use long division to simplify
the integral. To integrate the remaining rational expression …
a. Factor the denominator to a product of linear and quadratic polynomials.
b. For each linear factor (ax+b)m you get m fractions:
A
B
C

 ... 
2
ax  b (ax  b)
(ax  b) m
c. For each quadratic factor (ax2+bx+c)n you get n fractions:
Ax  B
Cx  D
Ex  F

 ... 
2
2
2
2
ax  bx  c (ax  bx  c)
(ax  bx  c) n
d. Set the original rational expression equal to the sum of the fractions from steps c & d and solve for
the constants.
e. Integrate each fraction.
7.
Numerical Integration
a. Midpoint Rule:

b
a
n
f ( x)dx  x f ( xk ) where x 
EM 
k 1
 a  x 2 k  1
ba
and xk  
n
 xk 1  x k  1
(b  a)3 K 2
where K 2  max f ''( x) over [a, b]
24n 2
b. Trapezoid Rule:

b
a
f ( x)dx 
ET 
n 1
x 
ba

f
(
a
)

2
f ( xk )  f (b)  where x 
and xk  a  k x


2 
n
k 1

(b  a)3 K 2
where K 2  max f ''( x) over [a, b]
12n2
c. Simpson’s Rule:
2
2 1

x 
ba
f ( x)dx 
f
(
a
)

4
f
(
x
)

2
f ( x2 k )  f (b)  where x 
and xk  a  k x



2 k 1
3 
n
k 1
k 1

n

b
a
n
(b  a)5 K 4
ES 
where K 4  max f (4) ( x) over [a, b]
4
180n
8.
Improper Integrals
b
k
a. If f(x) is continuous over [a,b), then  f ( x)dx  lim  f ( x)dx
k b
a
b
a
b
b. If f(x) is continuous over (a,b], then  f ( x)dx  lim  f ( x)dx
k a
a
b
k
c
k
c. If f(x) is continuous over (a,b), then  f ( x)dx  lim  f ( x)dx  lim  f ( x)dx , where c(a,b).
k a
a
k b
k
c
b
k
b
d. If f(x) is continuous over [a,b] except at c(a,b), then  f ( x)dx  lim  f ( x)dx  lim  f ( x)dx
k c
a

k c
a
k
k
e. If f(x) is continuous over [a,), then  f ( x)dx  lim  f ( x)dx
k  a
a
f. If f(x) is continuous over (-,b], then 
b

g. If f(x) is continuous over (-,), then 


h. And likewise for any other variation.
b
f ( x)dx  lim

f ( x)dx  lim

k  k
c
k  k
f ( x)dx
k
f ( x)dx  lim  f ( x)dx , where c(-,).
k  c