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Integration Rules and Techniques
Antiderivatives of Basic Functions
Power Rule (Complete)
xn^ dx =
xn+ n + 1
ln |x| + C, if n = − 1
Exponential Functions
With base a: (^) ∫
ax^ dx =
ax ln(a)
+ C
With base e, this becomes: (^) ∫
ex^ dx = ex^ + C
If we have base e and a linear function in the exponent, then ∫ eax+b^ dx =
a eax+b^ + C
Trigonometric Functions
sin(x) dx = − cos(x) + C ∫ sec^2 (x) dx = tan(x) + C ∫ sec(x) tan(x) dx = sec(x) + C
cos(x) dx = sin(x) + C ∫ csc^2 (x) dx = − cot(x) + C ∫ csc(x) cot(x) dx = − csc(x) + C
Inverse Trigonometric Functions
1 − x^2
dx = arcsin(x) + C ∫ 1 x
x^2 − 1
dx = arcsec(x) + C ∫ 1 1 + x^2
dx = arctan(x) + C
More generally,
∫ 1 a^2 + x^2
dx =
a
arctan
( (^) x a
+ C
Hyperbolic Functions
sinh(x) dx = cosh(x) + C ∫ cosh(x) dx = sinh(x) + C ∫ sech^2 (x) dx = tanh(x) + C
− csch(x) coth(x) dx = csch(x) + C ∫ − sech(x) tanh(x) dx = sech(x) + C ∫ − csch^2 (x) dx = coth(x) + C
Integration Theorems and Techniques
u-Substitution
If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then
∫ f (g(x)) g′(x) dx =
f (u) du
If we have a definite integral, then we can either change back to xs at the end and evaluate as usual; alternatively, we can leave the anti-derivative in terms of u, convert the limits of integration to us, and evaluate everything in terms of u without changing back to xs:
∫^ b
a
f (g(x)) g′(x) dx =
∫^ g(b)
g(a)
f (u) du
Integration by Parts
Recall the Product Rule: d dx
[u(x)v(x)] = v(x) du dx
Trigonometric Substitutions
With certain integrals we can use right triangles to help us determine a helpful substitution:
If the integral contains an ...then make the ...and... expression of the form... substitution... √ a^2 − x^2 x = a sin θ dx = a cos(θ) dθ
a^2 + x^2 x = a tan θ dx = a sec^2 (θ) dθ
x^2 − a^2 x = a sec θ dx = a sec(θ) tan(θ) dθ
You can memorize these rules if you wish, but we can also figure them out using a right triangle.
Partial Fraction Decomposition
Given a rational function to integrate, follow these steps:
- If the degree of the numerator is greater than or equal to that of the denominator perform long division.
- Factor the denominator into unique linear factors or irreducible quadratics.
- Split the rational function into a sum of partial fractions with unknown constants on top as follows:
A ︸ ︷︷ ︸ax^ +^ b for a linear factor
B
cx + d
C
(cx + d)^2
for a repeated linear factor
Dx + E ex^2 + f x + g ︸ ︷︷ ︸ for an irreducible quadratic
For example:
x^2 + 7 (2x + 1)(x − 3)^3 (x^2 + 3x + 1)^2
A
2 x + 1
B
x − 3
C
(x − 3)^2
D
(x − 3)^3
Ex + F x^2 + 3x + 1
Gx + H (x^2 + 3x + 1)^2
- Multiply both sides by the entire denominator and simplify.
- Solve for the unknown constants by using a system of equations or picking appropriate numbers to substitute in for x.
- Integrate each partial fraction. You may need to use u-substitution and/or
x^2 + a^2
dx =
a
tan−^1
( (^) x a
+ C.
Integration Using Tables
While computer algebra systems such as Mathematica have reduced the need for integration tables, sometimes the tables give a nicer or more useful form of the answer than the one that the CAS will yield. Oftentimes we will need to do some algebra or use u-substitution to get our integral to match an entry in the tables.
What follows is a selection of entries from the integration tables in Stewart’s Calculus, 7e:
a^2 + u^2 du =
u 2
a^2 + u^2 +
a^2 2
ln
u +
a^2 + u^2
+ C
u^2
a^2 + u^2 du =
u 8
(a^2 + 2u^2 )
a^2 + u^2 −
a^4 8
ln
u +
a^2 + u^2
+ C
a^2 + u^2 u
du =
a^2 + u^2 − a ln
a +
a^2 + u^2 u
+ C
a^2 + u^2 u^2
du = −
a^2 + u^2 u
u +
a^2 + u^2
+ C
du √ a^2 + u^2
= ln
u +
a^2 + u^2
+ C
u^2 du √ a^2 + u^2
u 2
a^2 + u^2 −
a^2 2 ln
u +
a^2 + u^2
+ C
du u
a^2 + u^2
a
ln
a^2 + u^2 + a u
+ C
Approximating Definite Integrals
We can approximate the net area under the graph of f (x) over the interval [a, b] using n rectangles and the indicated end/mid-point, where
∆x = b − a n
; x 0 = a, xi = a + i · ∆x, xn = b; xi = xi− 1 + xi 2
(the midpoint of the interval [xi− 1 , xi])
Left Endpoint Approximation: Uses the left endpoint of the subinterval to find the height of the corre- sponding rectangle.
∫^ b
a
f (x) dx ≈ Ln = ∆xf (x 0 ) + ∆xf (x 1 ) +... + ∆xf (xn− 1 )
= ∆x [f (x 0 ) + f (x 1 ) +... + f (xn− 1 )]
n∑− 1
i=
f (xi)∆x
Right Endpoint Approximation: Uses the right endpoint of the subinterval to find the height of the corresponding rectangle.
∫^ b
a
f (x) dx ≈ Rn = ∆xf (x 1 ) + ∆xf (x 2 ) +... + ∆xf (xn)
= ∆x [f (x 1 ) + f (x 2 ) +... + f (xn)]
∑^ n
i=
f (xi)∆x
