Download Daily Update: Week 2

Daily Update: Week 2
Math 146: Honors Calculus II, Spring 2016
Monday, January 25. We began class by Rfinding (or at least doing the majority of solving
x +1
dx, where we used the substitution u =
for) two antiderivatives. The first was 24x +1
2x . After substituting, we obtained a fairly simple rational function in u; we will need a
further method of integration (called “partial fractions decomposition”) to solve it.
R
The second was ln(ln(x))
dx, where we used the substitution Ru = ln(x), and then we
x
needed to use integration by parts to find the simplified integral ln(u) du!
For our lesson today, we discussed the first focus of §5.7, “Additional techniques of
integration,” which is integration of trigonometric functions. In particular, we focused on
(products of) powers of basic trig functions.R
We started with the question
of finding sin3 θ dθ. After applying the identity sin2 θ +
R
cos2 θ = 1, the integral became (1 − cos2 θ) sin θ dθ, and the substitution u = cos θ allowed
us to then find the antiderivative.
R
We noticed that a similar method works for sin4 x cos5 x dx, after rewriting the integrand as sin4 x(1 − sin2 x)2 cos x, so that the substitution u = cos x simplifies the integrand
enough to find the antiderivative. In fact, this method appears to work to find any
R
n
sin x cosm dx as long
(at least) one of the powers is odd.
R as
2
We noticed that sec
x
tan x dx could be solved with the substitution u = tan x. Then
R
4
we turned to finding
sec x tan x dx. After using the identity sec2 x = tan2 x + 1, our
R
2
2
integral became sec x(tan
1) dx, and the same substitution works. We noticed that
R x+
this method works to find secn x tanm x dx whenever the power of sec x is
R even.
What about the case when the power of sec x is odd? We approached tan3 θ sec θ dθ.
A substitution of u = tan θ does not appear to work, since even after using a trig identity
the power of sec x can never be two (aiming for “du = sec2 x dx”). Instead, we aimed to
use the substitution u = sec x, so that du = sec x tan x dx. This worked after rewriting the
integrand as follows:
tan3 θ sec θ = (sec2 θ − 1) tan θ sec θ.
This method appears to work whenever both
R the3 powers of sec x and of tan x are odd.
What about something analogous, like cot θ csc θ dθ? Since the derivatives of cot x
and csc x are analogous to tan x and sec x (at least up to a sign), and there is a trig identity
involving cot2 θ and csc2 θ we can use an analogous method of using a trig identity and
then a substitution.
R
We next turned to finding sin2 x dx, one of the “simplest” integral of the form
Z
sinn x cosm x dx
where both powers are even. After some reminders, we recalled that we used integration
by parts, along with the identity sin2 x + cos2 x = 1, to find this antiderivative. Instead,
we can use the double angle identity sin2 x = 21 (1 − cos(2x)), since the antiderivative of
this function can easily be used using the substitution u = 2x. Similarly, we can find
Daily Update: Week 2
2
R
sinm x dx or cosn x dx using this method if the power is even, by plugging in the appropriate double angle identity and simplifying. (Note that we probably will need to
recursively find an antiderivative with a smaller power in this process!)
Finally,
we turned to some “nontraditional” trig antiderivatives. The indefinite inteR
2
gral tan x dxRcould be solved immediately after substituting using the identity sec2 x =
tan2 x + 1, and tanR4 x dx could also be solved this way, except we needed to use a previous method to find sec4 x dxRalong the way.
Finally, we tried solving sec3 x dx using integration by parts, with u = sec x and
0
v = sec2 x dx. This gave us:
Z
Z
3
sec x dx = sec x tan x − sec x tan2 x dx.
R
As a challenge, think about how to solve this final integral!
Tuesday, January
26. With Leonard, we first finished calculating the “challenge” anR
3
tiderivative sec x dx from last time. Next, we found the following antiderivatives:
R
1. sec2 x csc2 x dx (using two identities works)
R
2. cot2 x csc4 x dx (substitution u = cot x works)
√
R
3. cos θ sin5 θ (substitution u = sin x works)
R 3x
4. cos
dx (substitution u = cos x works)
sin2 x
Wednesday, January 27. We started by giving a hint on one of the homework problems
in §5.7.
Next, we moved to finding the antiderivatives of tan(x) and sec(x); the antiderivatives
of cot(x) and csc(x) can be found analogously, respectively. Indeed, if these are needed on
a quiz or exam, you will either be given them, or specifically asked to derive the identity
yourself.
Next, we started the theory of integration via trigonometric substitution by attacking
the problem of showing that the area of the unit circle is exactly pi using calculus. We
R1√
shaded one quarter of the circle x2 +y 2 = 1, and equated it with the integral 0 1 − x2 dx.
p
√
Since 1 − sin2 θ = cos2 θ can be simplified to cos θ as long as cos θ ≥ 0 (e.g., if − π2 ≤
θ ≤ π2 , the integrand is simplified if we use the “reverse substitution” (i.e., x is written as
a function of the new variable θ instead of vice versa) x = sin θ:
Z √
Z p
Z
2
2
1 − x dx =
1 − sin θ cos θ dθ = cos2 θ dθ
as long as we restrict the values of θ to lie in the interval [− π2 , π2 ]. We used a double angle
identity to find this antiderivative, and then after changing the endpoints of the integral
Daily Update: Week 2
3
from x = 0 to x = 1 into θ = 0 to θ = π/2, we found that the area of a quarter circle is in
fact π/4, so that the area of the unit circle is, indeed, π.
2
We then turned to the problem of finding the area of the ellipse given by x4 + y 2 = 1,
which involved finding the antiderivative
Z r
Z
x2
1 √
1−
dx =
4 − x2 dx.
4
2
We noticed that the substitution x = sin θ would not get rid of the square roots as it did
for the circle; instead,
√ the substitution x = 2 sin θ works.
In general, if a2 − x2 appears in an integrand, where a is a constant, then the “reverse” trigonometric substitution
x = a sin θ can sometimes be helpful.
R √1−x2
dx using this method. After substitution and simplification
We next found
x
using techniques from last time, our indefinite integral became
Z
(csc θ − sin θ) dθ = − ln | cot θ + csc θ| + cos θ + C.
To return to the variable “x”, we drew a right triangle with angle opposite to θ of length
x
√ and hypotenuse of length one, since sin θ = x/1. Then the adjacent side has length
1 − x2 , and our integral became
√
√
√
1 − x2 1 √
+ + 1 − x2 + C = − ln | 1 − x2 + 1| − ln |x| + 2 − x2 + C.
− ln x
x
2
2
We noticed
substitu√ tan θ + 1 = sec θ gives two useful trigonometric
√ that the identity
π
2
2
tions: Since tan θ + 1 = sec θ = sec θ as long as sec θ ≥ 0 (for example, if − 2 < θ < π2 ),
√
then the substitution x = tan θ can√be useful if “ x2 + 1” appears in an integrand.
√
Similarly, since sec2 θ − 1 = tan2 θ = tan θ as long as tan θ ≥ 0 (for example, when
0 ≤ θ < π2 ), the substitution x = sec θ can be useful in finding the antiderivative of a
√
function involving R“ x2 − 1.”
We then found x√dx
dx using the substitution x = sec θ under the restriction that
x2 −1
π
0≤θ<
R 2 . After our substitution x = sec θ and dx = sec θ tan θ dθ, our integrand simplified to dθ = θ + C = arcsec θ + C, where the last equality holds because of our restriction
on θ.
Thursday,
January
R dx
R dx 28. In workshop, first we found formulas for the indefinite integrals
√
and 1+x2 (which we already know!) using the method of (reverse) trigonometric
1−x2
R
substitution. Next, we went on to a more complicated one, x2 √14−x2 dx, using the substitution x = 2 sin θ; our initial answer (in terms of θ) was − 41 cot θ + C, and in order to write
this in terms of the variable x, we drew a right triangle with one angle labeled as θ, and
the side opposite to θ with length x and with hypotenuse of length 2; we found
that (after
√
4−x2
using the Pythagorean theorem to solve for the final side’s length) cot θ = x .
4
Daily Update: Week 2
R
We then turned to using another way to find sec θ dθ. In case you want to re-create
θ
this method, try the following: multiply the integrand by 1 = cos
(and note that we
cos θ
would have to be careful about values of θ in general!), multiply products out, use a
trigonometric identity, and then substitute u = sin θ. From here, we needed a method
of “partial fractions decomposition;” Leonard explained the basic idea, and we solved
for a formula for the appropriate antiderivative. We will talk more about this method
tomorrow in lecture.
Finally, we had our first quiz.
Friday, January 29. Today we started class by finding the antiderivatives of several fairly
simple rational functions,
which are quotients of polynomial functions. For example,
R
using the fact that x21+1 dx = arctan(x) + C, we used the substitution u = x/2 to find that
R 1
dx = 14 arctan x4 + C; in general,
x2 +4
Z
x
1
1
dx
=
arctan
+ C.
x 2 + a2
a
a
R x
1
1
2
2
We found that 1+x
2 dx = 2 ln |1 + x | + C = 2 ln(1 + x ) + C (the last equality since
R
x2 + 1 > 0 for any value of x) using the basic substitution u = 1 + x2 . Turning to 5x+3
dx,
1+x2
5
2
we were able to find a full antiderivative as 2 ln(1 + x ) + 3 arctan(x) + C by splitting the
5
3
integrand into the sum of the two terms 1+x
2 and 1+x2 .
R 2 −2
dx can be found by rewriting the integrand as the sum of the
We noted that xx2 +1
x2 +2
3
terms x2 +2 = 1 and − x2 +1 . This, in fact, is the same result as we would get by doing
polynomial long division of x2 − 2 by x2 + 1 after writing the result as quotient plus
remainder divided by denominator; we did this same method to rewrite the following
rational function, allowing us to find its antiderivative using methods developed so far!:
−3x
x3 + 2x2 + 1
=x+2+ 2
.
2
x +1
x +1
Rewriting the integrand can be powerful in our attempts to find antiderivatives, as in
R
R sin3 x
examples like ln(x2 ) dx (using the fact that ln(x2 ) = 2x) and the quiz problem sec
5 x dx
5x+3
1
5
(using sec5 x = cos x). In this vein, we turned to finding a formula for x2 −1 dx; our methods
so far do not give a solution.
5x+3
A
B
It is a fact that the rational function 5x+3
= (x−1)(x+1
can be written as x−1
+ x+1
, where
x2 −1
A and B are real numbers; an analogous statement is true for the quotient any linear (or
constant) function by the product of two distinct linear functions. This rewriting is called
“partial fractions decomposition,” and we will see that we can similarly rewrite more
general rational functions later on in lecture.
We equated the two functions in question, and multiplied the equation through by the
left-hand-side’s denominator, obtaining
5x + 3 − A(x + 1) + B(x − 1).
Daily Update: Week 2
5
By plugging in x = 1 into the equation (one root of the denominator), we saw that 8 = 2A,
so that A = 4. Similarly, by plugging x = −1 in, we saw that −2 = −2B, so B = 1. Thus,
4
1
5x+3
= x−1
+ x+1
, and its antiderivative is 4 ln |x − 1| + ln |x + 1| + C. This
we have that (x−1)(x+1
far different than the result we got when the denominator was x2 + 1!
We noted that this method was used yesterday to find ∈ sec x dx.
We gave another example of rewriting an integrand in this way:
−11x + 3
−11x + 3
A
B
=
=
+
.
2
2x + x − 1
(2x − 1)(x + 1
2x − 1 x + 1
3
.
We found that A = 53 and B = − 14
2x3 −11x2 −2x+2
For something like
, we have to first do polynomial long division, so that
2x2 +x+1
the degree of the polynomial in the numerator is less than two, the degree of the denominator, and then we can apply the partial fractions decomposition method.
2 +100x−55
, since the denominator factors as x(x − 5)(x + 2), we can
For something like 3x
x3 −3x2 −10x
A
B
C
rewrite the rational function as x + x−5
+ x+2
.
However, it has mattered so far that the roots of the linear functions in the denominator of our rational functions are distinct (no repeats). If there are repeats, we are ensured
that the rational function can be instead rewritten in a slightly different way.
A
B
C
x2 +x+3
For example, (x−5)(x+2)
2 can be rewritten as a sum of the form x−5 + x+2 + (x+2)2 , where
A, B, and C are real numbers. To grasp the general form, if the denominator were instead
D
(x − 5)(x + 2)3 , we would also have to add a term of the form (x+2)
3 on to the aforementioned sum; generally, if a linear term has multiplicity n, there are n corresponding terms
in the sum, all with real number numerators. We found A, B, and C using our original
method, plugging in x = 5 and x = −2 first, and then using x = 0 on a whim, which
worked to find the third unknown real number.
2
. Its denominator can be factored as x(x2 + 4), but x2 + 4 canWe turned to 2xx3−x+4
+4x
not be factored into the product of two linear functions. (We can see this by solving for
roots/checking the discriminant.) In this case, the partial fractions decomposition has the
form
A Bx + C
2x2 − x + 4
= + 2
.
3
x + 4x
x
x +4
Notice the big difference here is that the numerator of x2 +4 may not be just a real number,
but a linear function! We solved for A, B, and C using a different method: after multiplying the above equation through by the left-hand side’s denominator, we equated the
coefficients of x2 and x on either side, as well as the constant terms.
We noticed throughout that we can, in fact, find the antiderivatives of each partial
fractions decomposition of the functions in question.
Finally, we turned to a rational function with x2 + 2x + 4 in the denominator, which
like x2 + 4, cannot be factored. However, we cannot simply equate
Rwe determined,
1
dx immediately with a function involving arctangent. Students suggested first
x2 +2x+4
completing the square of the denominator. In fact, this works! After this, we can do a
substitution, and then finding the antiderivative (and arctangent appears!).