Download Integration Techniques

Integration Techniques
Dr. Ebrahimian
Throughout this document n, m1 , . . . , mn are nonnegative constant integers and r, s, a, a1 , . . . , an , b1 , . . . , bn
are real constants.
1
1.1
Standard Substitutions
Denominator and Radical Substitutions
Consider substituting for a denominator or an expression under a radical.
x2
dx
x+1
R
5x
Example 2. √
dx
2x + 3
R
dx
√
Example 3.
(1 + x)4
Example 1.
1.2
R
R
f (u)u0 dx
Consider substituting for an expression that its derivative is a part of the integrand.
Example 4.
R
cos2 x sin x dx
Example 5.
R
(2x + 1)ex
2
+x
dx
2
R (ln x)
dx
x
√
R sin x
√
Example 7.
dx
x
R
Example 8. x−2 sec2 ( x1 ) dx
R
x
Example 9.
dx
1 + x4
Example 6.
1.3
Complete the Square
1
dx
x2 + 2x + 3
R
x
Example 11.
dx
4
x + 2x2 + 2
Example 10.
2
2.1
R
Integration by parts
Common Forms
Consider using integration by parts if you can break the integrand into two parts, one of which can be integrated and the other has a simple derivative.
1
If the integrand is a polynomial times another function, consider using the polynomial part as “u” in the
R
R
integration by parts formula (i.e. u dv = uv − v du) to reduce the degree of that polynomial.
If one part of the integrand is ln x consider using u = ln x since the derivative of ln x is simpler than its integral.
Example 12.
R
xex dx
Example 13.
R
x3 cos x dx
Example 14.
R
ln x dx
tan−1 x dx
R √
Example 16. ln 5 x dx
R 3
Example 17. x 4 ln x dx
Example 15.
2.2
R
Persist
Sometimes using integration by parts multiple times could help, even if it seems it doesn’t get you anywhere!
Example 18.
R
ex cos x dx
Example 19. Find a recursive formula for
3
sinn x dx.
Partial Fractions
In this section we consider integrals of form
3.1
R
R P (x)
dx where P (x) and Q(x) are polynomials.
Q(x)
Divide
If the degree of the numerator is more than or equal to the degree of the denominator consider dividing!
Before trying anything else it helps to perform the division.
R x3 + x + 1
dx
x2 + 1
R cos2 x sin x + 2 sin x
Example 21.
dx
cos2 x + 1
Example 20.
3.2
All Roots of Q(x) are Simple and Real
Q
P (x) P bi
(x − ai ) write
=
and
Q(x)
x − ai
find constants bi by multiplying both sides by Q(x) and then substituting x = ai .
If the denominator can be factored into distinct linear factors Q(x) =
1
dx
−x−2
R x3
Example 23.
dx
x2 − 1
Example 22.
R
x2
2
3.3
All Roots of Q(x) are Real but Some are Repeated.
If Q(x) =
n
Q
(x − ai )mi then for each terms (x − ai )mi we consider mi rational functions. We write
i=1
P (x)
as
Q(x)
r
where j ranges from 1 to mi . Then find constants r by multiplying both sides by Q(x)
(x − ai )j
and replacing x = ai , then repeatedly differentiationg both sides and replacing x = ai .
a sum of
Example 24.
R
Example 25.
R
3.4
x3
x+1
dx
− x2 − x + 1
x5
dx
x3 + 3x2 + 3x + 1
Q(x) has Quadratic Factors
When Q(x) has quadratic factors, that cannot be factored over reals, for each quadratic factor ax2 + bx + c
rx + s
consider a fraction like
. If Q(x) has repeated quadratic factors consider fractions of form
ax2 + bx + c
rx + s
where n ranges from 1 to the number of factors of ax2 + bx + c in Q(x).
(ax2 + bx + c)n
Example 26.
R
Example 27.
R
Example 28.
R
3.5
1
dx
x4 − 1
e2x
e2x
dx
+ 4ex + 4
1
dx
x4 + 4
Rationalizing Substitutions
Sometimes it is useful to substitute for the term involving radical and turn the integrand into a rational
expression, then use partial fractions.
√
1+x
dx
x
√
R
x
Example 30.
dx
x2 + x
Example 29.
4
4.1
R
Trigonometric Functions
R
cosr x · sins x dx
If r is an odd integer consider the substitution u = sin x. Then break cosr x into cosr−1 x · cos x. Next write
cosr−1 x in terms of sin x by noticing that cosr−1 x = (cos2 x)
R
r−1
into (1 − u2 ) 2 us du.
r−1
2
= (1 − sin2 x)
Similarly if s is an odd integer consider the substitution u = cos x.
3
r−1
2
. That turns the integral
If r and s are both even, use double angle formulas cos2 x =
the binomial formula and work on the resulting integrals.
Example 31.
R
sin3 x cos2 x dx
Example 32.
R
tan x dx
Example 33.
R
cos4 x dx
1 + cos 2x
1 − cos 2x
and sin2 x =
. Expand by
2
2
R sin4 x
dx
cos x
√
R sin3 ( x)
√
Example 35.
dx
x
Example 34.
Sometimes using some trig identities helps simplify the solution.
Example 36.
4.2
R
R
sin4 x cos4 x dx
tanr x · secs x dx
If r is an odd integer consider the substitution u = sec x. Write tanr x·secs x as tanr−1 x·secs−1 x·tan x·sec x.
Notice that the derivative of sec x is tan x·sec x and since r−1 is even you can write tanr−1 x = (sec2 x−1)
r−1
2
.
So the substitution u = sec x could help.
When s is even, use secs x = secs−2 x · sec2 x and notice that sec2 x is the derivative of tan x and secs−2 x =
(tan2 x + 1)
s−2
2
Example 37.
. So the substitution u = tan x could help.
R
tan3 x dx
R tan3 x
dx
cos3 x
If r is even and s is odd it might be helpful to express the integrand in terms of sec x. Then use integration
Example 38.
by parts to integrate powers of sec x.
Example 39.
4.3
R
R
sec3 x dx
sin rx · cos sx dx,
R
sin rx · sin sx dx and
Consider using the identities
• 2 cos x · cos y = cos(x + y) + cos(x − y)
• 2 sin x · sin y = cos(x − y) − cos(x + y)
• 2 sin x · cos y = sin(x + y) + sin(x − y)
R
Example 40. cos 3x cos 2x dx
R
Example 41. sin(x + 1) cos 2x dx
4
R
cos rx · cos sx dx
4.4
Rational Functions of sin2 x and cos2 x
If the integrand is a function of sin2 x and cos2 x consider using the identities sin2 x =
cos2 x =
1
. Then use u = tan x, and notice that du = (1 + u2 ) dx.
1 + tan2 x
Example 42.
4.5
R
tan2 x
and
1 + tan2 x
tan2 x
dx
1 + cos2 x
Rational Functions of cos x and sin x
When integrating a rational function in terms of sin x and cos x the substitution u = tan x2 turns the inte1 − u2
2u
grand into a rational function of u, because cos x =
and sin x =
. These trig identities can be
1 + u2
1 + u2
proved using double angle formulas.
Example 43.
R
1
dx
2 sin x − 3 cos x
Example 44.
R
1
dx
1 + sin x − cos x
5
Trigonometric Substitutions
When the expression
π
π
− <t< .
2
2
When the expression
π
π
− ≤t≤ .
2
2
√
x2 + a2 is a part of the integrand consider using the substitution x = a tan t where
√
a2 − x2 is a part of the integrand consider using the substitution x = a sin t where
√
When the expression x2 − a2 is a part of the integrand consider using the substitution x = a sec t where
π
3π
0 ≤ t < or π ≤ t <
.
2
2
Example 45.
R
Example 46.
R
√
√
x2
dx
3 − x2
1 − x2
dx
x2
1
dx
9 + x2
R
x
Example 48.
dx
4 − 2x − x2
Example 47.
6
6.1
R
x2
√
Definite Integrals
Symmetries
If f is an odd function then
Ra
−a
f (x) dx = 0.
5
Example 49.
R1
4
xex dx
−1
6.2
Switching the Limits
In general when evaluating a definite integral of form
Rb
f (x) dx consider the substitution u = b + a − x or
a
some other substitution that switches a and b. Adding the resulting integrals might help in evaluating the
original definite integral.
π
For example when evaluating
R2
f (sin x, cos x) dx, the u−substitution u =
0
π
2
− x would turn the definite
π
2
integral into
R
f (cos u, sin u) du. Then adding the two might help in evaluating the definite integral.
0
π
Example 50.
R2
0
sinn x
dx
sinn x + cosn x
π
Example 51.
0
Example 52.
7
1
R2
√
1 + (tan x)
2
dx
R∞ ln x
dx
2
0 1+x
Miscellaneous Examples
Example 53.
R
Example 54.
R
√
1
√ dx
x+ 3x
x2
1
dx
− 2x
R ln x
√ dx
x
p
√
R 1+ x
Example 56.
dx
x
R
Example 57. x sin3 x dx
Example 55.
Example 58.
R
x sec x tan x dx
Example 59.
R
dx
cos x − 1
Example 60.
R
1
dx
x7 − x
Example 61.
R
ecos x sin 2x dx
Example 62.
R
e
Example 63.
R√
√
x
dx
tan x dx
√
R 1−x
Example 64. √
dx
1+x
6
R sin x + sec x
tan x
5
R cos x
Example 66. √
dx
sin x
R
Example 67. (sin−1 x)2 dx
R
Example 68. sin(ln x) dx
Example 65.
3x2 + 1
dx
1 + 2x + + 2x3 + 2x4 + x6
Example 69.
R
Example 70.
R
1
dx
1 + cos2 x
Example 71.
R
√
x2
x+
1
√
1+x
dx
R ln(1 + x)
dx
x2
R
Example 73. xn ln x dx
R
Example 74. xn ex dx
R
Example 75. ln(x2 − x + 1) dx
Example 72.
1
dx
xn (x − a)
√
R p
Example 77. x 1 − 4 − x2 dx
R
Example 78. x tan2 x dx
Example 76.
R
Example 79.
R
Example 80.
R2
x5
1
dx
+x
π
sin2n+1 x dx
0
π
Example 81.
R2
sin2n x dx
0
Example 82. Use the above two examples to prove limn→∞
Example 83.
R1 √
3
1 − x7 −
√
7
1 − x3 dx
0
Example 84.
Example 85.
R∞ ln x
dx
2
0 1+x
R1
(ln x)n dx
0
Example 86.
R1
(1 − x2 )n dx
0
Example 87.
R∞
−1
x8
dx
(1 + x6 )2
7
24n (n!)4
π
= .
((2n)!)2 (2n + 1)
2