Download Integration techniques

Integration techniques
Basic integrals
∫
∫
1
xn dx = n+1
xn+1 + C, n ̸= −1
∫
sec2 x dx = tan x + C
∫
csc x cot x dx = − csc x + C
∫ x
a dx = ln1a ax + C
√ 1
x2 −1
dx = arcsec x + C, x > 1
∫
∫
∫
∫
∫
sin x dx = − cos x + C
csc2 x dx = − cot x + C
∫ x
e dx = ex + C
√ 1
1−x2
dx = arcsin x + C
tan x dx = ln | sec x| + C
cos x dx = sin x + c
∫
∫
sec x tan x dx = sec x + C
∫ 1
dx = ln |x| + C, x ̸= 0
x
∫ 1
dx = arctan x + C
1+x2
sec x dx = ln | sec x + tan x| + C
The substitution rule
If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I:
∫
∫
′
f (g(x))g (x) dx =
f (u) du
If g ′ is continuous on [a, b] and f is continuous on the range of u = g(x), then
∫b
∫g(b)
f (g(x))g ′ (x) dx =
f (u) du
a
g(a)
Integration by parts
∫
∫
′
f (x)g(x) dx = f (x)g(x) −
∫b
∫b
′
f (x)g(x) dx =
a
f (x)g ′ (x) dx
f (x)g(x)|ba
∫
∫
u dv = uv −
−
f (x)g ′ (x) dx
a
v du
Recall LIATE: log - inverse trig - algebraic (polynomial and rational) - trig - exponential.
Usually, what comes first in the acronym is what you differentiate, i.e. g(x).
Trigonometric integrals
Integral of the type
∫
sinm (x) cosn (x) dx:
1
a) If n = 2k + 1 (odd), save one cosine factor, use cos2 x = 1 − sin2 x and substitute
u = sin x.
b) If m = 2k+1 (odd), save one sine factor, use sin2 x = 1−cos2 x and substitute u = cos x.
c) If both powers are even, use the half angle identities: sin2 x = 21 (1 − cos(2x)), cos2 x =
1
(1
2
+ cos(2x)) and 2 sin x cos x = sin(2x).
∫
Integral of the type tanm (x) secn (x) dx:
a) If n = 2k (even), save a factor of sec2 (x), use sec2 (x) = 1 + tan2 (x) and substitute
u = tan(x).
b) If m = 2k + 1 (odd), save a factor of sec(x) tan(x), use tan2 (x) = sec2 (x) − 1 and
substitute u = sec(x).
b) If m = 2k (even) and n = 2j + 1 (odd), use tan2 x = sec2 x − 1 to express tanm x in
terms of sec x. Then use integration by parts to integrate odd powers of sec x.
d) For any other case, write the expression in terms of sin(x) and cos(x).
Integrating products of sines and cosines of different angles
Use the following substitutions:
1
1
cos[(a − b)x] − cos[(a + b)x]
2
2
1
1
sin(ax) cos(bx) = sin[(a − b)x] + sin[(a + b)x]
2
2
1
1
cos(ax) cos(bx) = cos[(a − b)x] + cos[(a + b)x]
2
2
sin(ax) sin(bx) =
Trigonometric substitutions
Expression
√
a2 − x2
√
a2 + x2
√
x2 − a2
Substitution
Differential
Identity
x = a sin t, − π2 ≤ t ≤
π
2
dx = a cos t dt
1 − sin2 t = cos2 t
x = a tan t, − π2 < t <
π
2
dx = a sec2 t dt
1 + tan2 t = sec2 t
dx = a sec t tan t dt
sec2 t − 1 = tan2 t
x = a sec t, 0 ≤ t ≤
3π
,
2
t ̸=
π
2
Using partial fraction decomposition
∫ (x)
Rational functions integrals fg(x)
dx need to be written as a sum of partial fractions.
The degree of f (x) must be less than the degree of g(x). If it isn’t, first divide f (x) by g(x)
(i.e. long division).
2
Any polynomial with real coefficients can be written as a product of real linear factors and
real quadratic factors. If the factors of g(x) are known, proceed as follows:
1. Suppose (x − r)n is one of the factors of g. To this factor assign the sum:
A1
A2
A3
An
+
+
+ ... +
2
3
(x − r) (x − r)
(x − r)
(x − r)n
2. Suppose (x2 + px + q)n is one of the factors of g, such that x2 + px + q has no real roots.
To this factor, assign the sum of n partial fractions:
B3 x + C3
Bn x + Cn
B1 x + C1
B2 x + C2
+
+
...
+
+
(x2 + px + q) (x2 + px + q)2 (x2 + px + q)3
(x2 + px + q)n
3. Set the original fraction equal to the sum of all these partial fractions. Clear the resulting
equation of fractions and arrange the terms in decreasing order of the powers of x.
4. Equate the coefficients of corresponding powers of x and solve the resulting equations
for the undetermined coefficients. One may use the cover method sometimes, as well as
considering random values for x and obtaining as many equations as needed.
Integrating the basic factors in the decomposition
2
2
The irreducible terms x2 + px + q can be written as x2 + px + q = x2 + 2 12 px + p4 + q − p4 =
)2
(
2
x + p2 + q − p4 = y 2 + r2 so make the substitution y = x + p2 .
There are 6 basic terms:
∫
1
x−a
∫
1
(x − a)n
∫
1
2
x + a2
∫
x
x2 + a2
∫
x
2
(x + a2 )n
∫
1
2
(x + a2 )n
dx = ln |x − a| + C
1
1
·
+C
n − 1 (x − a)n−1
(x)
1
dx = arctan
+C
a
a
1
dx = ln(x2 + a2 ) + C
2
1
1
dx = −
·
+C
2(n − 1) (x − a)n−1
∫
1
dx = 2n−1 [cos(θ)]2n−2 dθ
a
dx = −
where for the last integral one uses the substitution x = a tan θ. The resulting integral is solved
using the rules from trigonometric integrals.
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