Unformatted text preview: ring the coefﬁcients of , and get a system of equations. Alternatively, we may
clear the denominators, then use particular values for to create systems of equations; this
method is extremely useful when using roots of to eliminate terms. 6 WATERLOO SOS E XAM AID: MATH138 M IDTERM
Step 4: Integrate term by term from step 3. Recall that
if
if
For the quadratic case, we ﬁrst complete the square. Once we are in the form
we can break up the integral as
The ﬁrst integral is simple, since differentiating gives a term, so
The second integral can be completed using a trigonometric substitution , with
:
which we can solve using previous techniques. Notice that in the case , this reduces to
as expected.
Rationalizing Substitution:
If the integrand contains a term of the form 3 Strategy for Integration (7.5)
, then the substitution may be effective. The textbook uses a fourstep strategy to solve integrals, which is very useful, but not perfect  integration
is hard!
(1) Simplify the integrand if possible.
(2) Look for an obvious substitution or antiderivative.
(3) Attempt to use one of the ﬁve methods you have learned:
(i) Integration by parts
(ii) Trigonometric integration
(iii) Trigonometric substitution
(iv) Partial fractions
(v) Rationalizing substitution
7 WATERLOO SOS E XAM AID: MATH138 M IDTERM
(4) Try again. E XAMPLE 1. Calculate
S OLUTION. The strategy is not immediately obvious. There is no clear simpliﬁcation to be made, no trigonometric functions to substitute, no radicals, and no rational functions. However, notice that the derivative of
will yield a term, so we try for a direct antiderivative. To get in the denominator,
we use a logarithm:
Taking the derivative, we see by chain rule that
Thus, to get to our integrand, we require a negative sign:
E XAMPLE 2. Calculate S OLUTION. We have a product of functions, so we try integration by parts:
Then
The integral is no easier than what we started with, but we can use integration by parts a second time, to
get a recurring integral:
Then
8 WATERLOO SOS E XAM AID: MATH138 M IDTERM
We solve for our original integral using simple algebra now.
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This document was uploaded on 03/04/2014 for the course MATH 138 at Waterloo.
 Winter '07
 Anoymous
 Math, Calculus

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