0.1 Substitution
When: Given a function f(x) inside a different function g(x), with f'(x) outside g(x).
Necessary: An integral 
.
How: Let u = f(x), du = f'(x)dx
.
Check: After substitution, there should no longer be any x's around.
Why? This is essentially the chain rule for differentiation in reverse.
1 Chapter 7: Techniques of Integration
1.1 7.1: Integ ration by Parts
When: Given two functions f(x), g'(x) that don't cooperate through integration
or differentiation.
Essentially, when substitution for f(x) or g'(x) won't work. Also requires that f(x) is
differentiable and g'(x) can be integrated.
Necessary: An integral 
.
How: Let u = f(x), du = f'(x)dx
Check: differentiate v to see that it gets dv.
Some integrations by parts takes several attempts to remove the integral part.
Some will
never actually remove the integral through normal integration. What is the trick
when
integrating 
?
Why? This is the product rule for differentiation in reverse, with a bit of
term-shifting.
1.2 7.2: Trigonometric Integrals
When: Given an integral in the form of powers and multiples of trigonometric
functions.
How: Use trigonometric identities. Big ones are :
See full list in Section 1.3, pp. 26-27.
Reduce the integral into mostly one function (For example, all in sin x) and a
single repre-
sentation of that function's derivative (cos x for this example). Now,
substitute u = sin x
and integrate.
Check: differentiate v to see that it gets dv.
Why? Trigonometric integrals are more complicated after using substitutions in
7.3.
1.3 7.3: Trigonometric Substitutions
When: The integral has 
, or
somewhere in the function. This entity
will be referred to as S from now on.
How: Consider a, x, and S as side lengths for a right triangle with an angleθ
(see Figure 7.2 on
p. 461). Use Pythagorean Theorem to determine which are legs and which is
hypotenuse.
Then, use trig functions to substitute in place of x, dx, a, and S.
After this new integral is solved, use the triangle diagram to replace
trigonometric functions
in θ with appropriate ratios of x, a, and S.
Check: differentiate, or check algebra and trig identities really well .
Why? Applying geometry to calculus recalls the importance of high school math .
1.4 7.4: Integration by Partial Fraction Decomposition
When: When given a rational function 
where
f(x) and g(x) are polynomials and g (x) factors.
How: First, the degree of f(x) must be less than degree of g(x). If not, use
long division of
polynomials to get 
where the degree of r(x)
is less than the degree of
g(x). Also, multiply out by a constant so the first term of g(x) is xd, or so the
first coefficient
is 1.
Now, split g(x) into linear and quadratic factors:
Set up an equation with on the left.
For each linear term (x - a)m in g(x), add 
to
the right.
For each quadratic term (x2 + bx + c)n in g(x), add
to the right.
Now, multiply by g(x) on both sides to get an equality of polynomials. By
combining like
terms on the right, we get a system of equations between the coefficients on the
left and the
coefficients on the right. Solve for the unknown A's, B's, and C's to result in
the partial
fraction expansion.
Integrate each partial fraction individually, and combine the resulting
antiderivatives.
Check: Add the expansion together by finding a common denominator. You should get
the original
rational function back. Also, combine logarithms! See properties of logarithms
on p . 41.
Why? Polynomials are easy, but rational functions are more difficult. Solve by
breaking the prob-
lem into several smaller problems.
Trick: Sometimes a substitution will turn a complicated function into a rational
function. See
problems 35-38 in Section 7.4, p. 470.
1.5 7.5: Integral Tables and Computer Algebra Systems
When: The integral is a familiar function, such as a polynomial, trigonometric
function, or the
derivative of a well-known function.
How: Memorize the integrals and derivatives that show up a lot. I recommend
knowing how to
integrate and differentiate all the trig functions.
To show your work, write 
to show
.
Be careful to make sure to have an exact substitution! If you have constant
multiples, make
sure they are cancelled properly.
Why? Certain integrals show up frequently near the end of big problems, and can
be solved quickly
to save time for reverse-substitution or doing other problems.
1.6 7.6: Numerical Integration
When: In order to approximate integrals that cannot be solved, we need a minimum
number of
intervals to use for computation in order to reach a certain error bound.
How: For the trapezoid rule, we need f''(x) to be continuous.
Let M be an upper bound for f''(x) for x ∈ [a, b].
Now, solve for n in the equation 
, where E is
the maximum allowed error.
For Simpson's rule, we need f(4)(x) (the fourth derivative of f(x)) to be
continuous.
Let M be an upper bound for f(4)(x) for x ∈[a, b].
Now, solve for n in the equation 
, where E is
the maximum allowed error.
Why? Not all functions have integrals, so a computer would be used to compute an
approximate
answer, but the number of intervals required can be predicted.
1.7 7.7: Improper Integrals
When: If given an integral with infinite bounds, or a definite integral where f(x)
is discontinuous
at or within the bounds.
How: Take the limit!
Type I.a: 
.
Type I.b: 
.
Type I.c: 
.
Type II.a: f(x) discontinuous at b, 
.
Type II.b: f(x) discontinuous at a, 
.
Type II.c: f(x) discontinuous at c ∈ (a, b),
.
Why? infinite regions may have finite area if a function converges to 0 fast
enough. Also, functions
may be point-wise discontinuous but still have useful integrals.
