Strategy for Integration by Parts
1.  xn f(x) dx where f(x) = sin ax, cos ax, eax (where a is a nonzero constant)
Let u = xn and dv = f(x) dx (use tabular method when n  2)
2.  xn f(x) dx where f(x) = ln ax, sin–1ax, tan–1ax, sec–1ax, (where a is a nonzero constant)
Let u = f(x) and dv = xn dx (including the case n = 0)
3.  eax sin (bx) dx and  eax cos (bx) dx (where a and b nonzero constants)
You have to use integration by parts (udv-substitution) twice. The first time udv-substitution is arbitrary,
for example,  eax sin (bx) dx, you can let u = eax and dv = sin (bx) dx or u = sin (bx) dx and dv = eax dx.
However, the second time udv-substitution must follow the same substitution you use for the first time.
Of course, the most important thing is, after you carry out the integration by parts twice, you will see the
integral  eax sin (bx) dx again (but with minus sign before it). What you need to do now is to transpose
this negative integral to the other side.
Note:
1. Here, we assume n (the power of x) is a nonnegative integer.
Strategy for Trigonometric Integrals
For  sinmx cosnx dx (m, n  0)
1. If m (the power of sine) is even and n (the power of cosine) is odd:
Let u = sin x, save one factor of cos x for du, and use cos2x = 1 – sin2x to express the remaining factors in
terms of sine. For example,
 sin4x cos3x dx =  sin4x cos2x cos x dx =  sin4x (1 – sin2x) cos x dx =  u4 (1 – u2) du = ...
Let u = sin x, du = cos x dx
2. If m (the power of sine) is odd and n (the power of cosine) is even:
Let u = cos x, save one factor of sin x for du, and use sin2x = 1 – cos2x to express the remaining factors in
terms of cosine. For example,
 sin5x cos2x dx =  sin4x cos2x sin x dx =  (1 – cos2x)2 cos2x sin x dx = – (1 – u2)2 u2 du = ...
Let u = cos x, du = –sin x dx
3. If m (the power of sine) is odd and n (the power of cosine) is odd:
Use either one of the above two methods. For example,
 sin3x cos3x dx =  sin3x cos2x cos x dx =  sin3x (1 – sin2x) cos x dx =  u3 (1 – u2) du = ...
Let u = sin x, du = cos x dx
Or
 sin3x cos3x dx =  sin2x cos3x cos x dx =  sin2x cos3x sin x dx = – (1 – u2) u3 du = ...
Let u = cos x, du = –sin x dx
4. If m (the power of sine) is even and n (the power of cosine) is even:
You have to use the identities: sin2x = ½(1 – cos 2x) and cos2x = ½(1 + cos 2x). For example,
 sin2x cos2x dx =  ½(1 – cos 2x)½(1 + cos 2x) dx = ¼ 1 – cos22x dx = ...
Notes:
1. If either or both powers of sine and cosine are odd, you may need to use one of the following identities:
cos2x = 1 – sin2x and sin2x = 1 – cos2x (both of these are from cos2x + sin2x = 1).
2. If both powers of sine and cosine are even, you must use the identities: sin2x = ½(1 – cos 2x) and cos2x =
½(1 + cos 2x).
3. It is also helpful for you to know the following identities: (a + b)(a – b) = a2 – b2, (a + b)2 = a2 + 2ab +
b2, and (a – b)2 = a2 – 2ab + b2.
For  tanmx secnx dx (m, n  0)
1. If n (the power of secant) is even:
Let u = tan x, save one factor of sec2x for du and use sec2x = 1 + tan2x to express the remaining factor of
secant in terms of tangent. For example,
 tan5x sec4x dx =  tan5x sec2x sec2x dx =  u5 (1 + u2) du =...
Let u = tan x, du = sec2x dx, and sec2x = 1 + tan2x = 1 + u2
2. If m (the power of tangent) is odd:
Let u = sec x, save one factor of tan x sec x for du and use tan2x = sec2x – 1 to express the remaining
factor of tangent in terms of secant. For example,
 tan5x sec4x dx =  tan4x sec3x tan x sec x dx =  (u2 – 1)2 u3 du =...
Let u = sec x, du = tan x sec x dx, and tan4x = (tan2x)2 = (sec2x – 1)2 = (u2 – 1)2
Summary:
Power of tangent
Odd
Odd
Even
Even
Power of secant
Odd
Even
Odd
Even
Method
2
1 or 2
Integration by Parts
1
Strategy for Trigonometric Substitution
Use trigonometric substitution when you see the integrand involves x 2  a 2 ,
there is no obvious u-substitution.
x 2  a 2 or
a 2  x 2 and
Draw a right triangle, let one of the acute angles (usually the one at the base) be ,
If the integrand has
x a
2
2
x2  a2
Let
hyp. = x  a
opp. = x, adj. = a
2
hyp. = x
opp. =
a2  x2
2
x 2  a 2 , adj. = a
hyp. = a
opp. = x, adj. =
a2  x2
Triangle should be drawn Trig. function used for dx
tan  = x/a
x2  a2
x
sec2 d = 1/a dx

a sec2 = dx
a
sec  = x/a
x
x2  a2
sec  tan  d = 1/a dx

a sec  tan  = dx
a
sin  = x/a
a
x
cos  d = 1/a dx

a cos = dx
x2  a2
Then break the integral into several components (usually 3—the numerator, the denominator, and dx), and
rewrite each component in terms of , using trigonometric functions (of course). For dx, always use the
trigonometric function that is = to x/a.
Example:
dx

x 2 16 x 2  9
4x

16 x 2  9

3
1
1
16 1 3
4

 dx     cos 2   cot   sec  tan  d   cos  d
2
x
9 3 4
9
16 x 2  9
4
3
4x
3
cos  
sec  
 sin   C
cot  
4x
3
9
16 x  9
2
4
3 4
(cos   )
3
4x 3
4
1
cos  
3
x
16
1
2
cos   2
9
x
1
3
1
(cot  
)
3
16 x 2  9 3
1
1
cos  
3
16 x 2  9
3
4x 3
(sec   )
4
3 4
3
sec   x
4
4 16 x 2  9
 
C
9
4x
3
sec  tan  d  dx
4

16 x 2  9
C
9x
Note:
1. Here, we don’t have the 3 usual components (i.e., the numerator, the denominator, and dx), but we still
can break it down to 3 components (2 for the denominator and 1 for dx).
2. Since the denominator has 16 x 2  9  (4 x) 2  32 , so the hypotenuse should be 4x instead of x.
3. Notice the second column (the “Let” column) of the table has the hypotenuse in bold, which means the
hypotenuse must be that quantity (or similar quantity). On the other hand, the quantities for the two legs
can be interchanged (here, I just give my preferred setup of the triangle).
Alternative Method:
If the integrand has
x2  a2
Let
x = a tan 
Identity should be used
1 + tan2 = sec2
dx (from the “Let” column)
dx = a sec2 d
x2  a2
x = a sec 
sec2 –1 = tan2
dx = a sec  tan  d
a2  x2
x = a sin 
1 – sin2 = cos2
dx = a cos d
Example (same as above):
3
dx
4 sec  tan  d

1
 2

( 34 sec  ) 2  [16( 34 sec  ) 2  9] 2
x 16 x 2  9
let 4x = 3sec 
x = ¾ sec 
dx = ¾ sec tan  d

SW for the denominator:
1
1
( 34 sec  ) 2  [16( 34 sec  ) 2  9] 2  169 sec 2   [16( 169 sec 2  )  9] 2
sec  tan  d
sec 2  tan 
 169 sec2   [9sec2   9] 2
3
4
27
16
1
 169 sec2   [9(sec2   1)] 2
1
3 16
1
  
d
4 27 sec 
4
  cos  d
9
 169 sec2   [9 tan 2  ] 2
1
27
 169 sec2   3 tan   16
sec2  tan 
4x
4
4 16 x  9
16 x  9
 sin   C  
C 
C
9
9
4x
9x
2
2
16 x 2  9

3
Notes:
1. As before, since the denominator has 16 x 2  9  (4 x) 2  32 , we let 4x = 3sec , instead of x = 3sec .
2. Before we actually integrate the integral, all we do is substituting the components in terms of  and then
simplifying the expression by using trigonometric identities and law of exponents. After we integrate the
integral, however, if the answer involves a trigonometric function that is not same as the “let,” we still
have to draw out the triangle (like the case above).
Strategy for Integration by Partial Fractions:
f ( x)
 g ( x) dx
1. If degree deg(f(x)) < deg(g(x)), factor the denominator, g(x), into the product of linear factors and
irreducible quadratic factors. Then set up f(x)/g(x) as the sum of the partial fractions with denominators in
these linear and quadratic factors. For example,
f ( x)
where deg(f(x)) < 6
6
x  x2
 x6 – x2 = x2(x4 – 1) = x2(x2 – 1)( x2 + 1) = x2(x – 1)(x + 1)( x2 + 1)
f ( x)
A B
C
D
Ex  F
 6
  2

 2
2
x x
x x
x 1 x 1 x 1
And solve for A, B, C, D, E and F (they are all real numbers)
2. If deg(f(x))  deg(g(x)), then divide f(x) by g(x) using long division (i.e., g ( x) f ( x) ). Say after long
division, we find out f(x) = g(x)q(x) + r(x), where q(x) is the quotient with degree  0 and r(x) is the
remainder with degree < f(x).
f ( x)
r ( x)
f ( x)
r ( x)
Notice that f(x) = g(x)q(x) + r(x) 
, so 
 q ( x) 
dx   q( x) 
dx and integrate
g ( x)
g ( x)
g ( x)
g ( x)
RHS. Of course, since the deg(r(x)) < deg(g(x)), you may need to decompose r(x)/g(x). For example,
x3  2 x
3x
A
B
 x 2  1 dx   x  x 2  1 dx   x  x  1  x  1 dx  ...
x
Note:
x 2  1 x3  2 x
1. Since the final answer of most of these integrals involves the ln function and
3
the tan–1 function, you will be better off by knowing the following:
x  x
k
k
k
3x
dx  k ln x  a  C
dx  ln bx  a  C (b  0)
a) 
b) 
xa
bx  a
b
k
k
x
kx
k
dx  tan 1  C
dx  ln  x 2  a 2   C
c)  2
d)  2
2
2
x a
a
a
x a
2
k
k
bx
kx
k
e) 
dx 
tan 1  C f) 
dx  2 ln  x 2  a 2   C
2
2
2
2
(bx)  a
ab
a
(bx)  a
2b