Calculus II
Class Notes, 1/29/99
In order to integrate cos(5z) * cos(12z), we could use integration
by parts in a fairly obvious but slightly complicated way. Instead,
for practice in dealing with situations where we can't immediately see how to find an
integral, we use the table in the back of the text.
Using #11 from the table with a = 5 and b =
12, with the formula indicated below, we can easily obtain an expression for the
integral by plugging in these values for a and b.
Similarly we use #14 for the integral of p(x) e^(ax) to
evaluate the integral in the figure below. Here p(x) = x^3 + 3 x^2 - 5,
and a = 2.
- The integral of p'(x) e^(ax) = (3 x^2 - 6x - 5) e^(2x) will
then require us to use the same formula again.
- Application of the formula to this new integral will lead to an integral of
6x - 6 e^(2x), to which we can apply the formula once more.
- Alternatively we can apply the second form of a
formula, which in this case will give us 1/a * p(x) e^(ax) - 1/a^2 * p'(x) e^(ax)
+ 1/a^3 * p''(x) e^(ax) - 1/a^4 * p'''(x) e^(ax). We note that p''''(x) =
0, so we need only use the four terms indicated.
We integrate sin^6(`theta) using the reduction formula,
#17, which reduces the power being integrated by
2.
- For n = 6 we obtain the result in the third line.
- We then integrate the fourth power of the sine
function using the same formula with n = 4.
- We finally integrate the square of the sine
function using either the n = 2 version of the formula or the formula we obtain
earlier.
We perform the integration below, with the irreducible quadratic in
the denominator, by first completing the square.
- We first note that the denominator cannot equal 0 (setting it equal to
0 using the quadratic formula we obtain a negative quantity under the square root) so we
cannot factor the denominator. This is important because if the denominator does factor,
we have a formula that we can use; and besides completing the square won't work the same
way if the denominator factors.
- Using standard techniques for completing the square, where we note that
x^2 + 8x is the beginning of the perfect square
(x + 4)^2 = x^2 + 8x + 16, we see that the denominator must be (x+4)^2
+ 14.
- Having represented the denominator in this form we see that the u substitution u
= x+4 immediately leads is to the final integral in the figure
below, which should suggest an inverse tangent. The actual
form is identical to that of #24 in the table.
- Using the integral given in the table, we obtain the integral in terms
of u. We then back substitute u = x+4 to get the
integral in terms of x.
When we have a polynomial divided by another polynomial, with the
polynomial in the numerator of greater degree than that
in the denominator, we often use a long division to
express the fraction in the form indicated below.
- When doing the long division, the main source of error is
careless subtraction. Be careful.
The integral of x-5 in the last line below is easily
found.
The fractional expression in the integral has a denominator which
factors, as indicated. The factored form leads us to an integral
whose form is that of #27 in the table, and which can therefore
be found by simple substitution.
The integral in the figure below is evaluated by using partial fractions.
- We know that when a fraction with denominator x-3 is
added to a fraction with denominator x+5 we will obtain
a fraction whose denominator is (x-3)(x+5).
- We conclude that must be possible to express the given fraction
as the sum A / (x-3) + B / (x+5), where A and B
are numbers to be determined.
- Setting the original fraction equal to the sum we
obtain an equation for A and B, as
expressed in the third line.
- Multiplying both sides of the equation by (x-3) (x+5) we
obtain the equation in the fourth line, which we rearrange to
the form in the fifth line by collecting the x terms and the constant
terms on the left-hand side and factoring.
- Since the right-hand side does not have an x
term, we see that A + B = 0; we see that therefore 5A -
3B = 1, so we have two equations in two unknowns A
and B.
- Solving these equations we obtain B = -1/8, A = 1/8,
as indicated.
- We conclude that the expression to be integrated is A / (x-3) + B / (x+5) = 1/8
* 1/(x-3) - 1/8 * 1/(x+5).
These results are represented in the integrals to the
right of the = sign in the first line.
These integrals are easily
evaluated by simple u substitution.
