1. Let u = cosx. Then du = s

MATH2007* Partial Answers to Review Exercises Fall 2004
Evaluate each of the following integrals:
1. Let u = cosx. Then du = − sin x dx and
Hence
sin 3 x dx = (1 − cos 2 x)(sin x dx) = (u 2 − 1) du.
3
sin x
cos7 2 u − 1
dx =
x u7 du = · · · · · · Please finish this.
2. We use integration by parts:
Please finish this.
(x 2 3 d x 5x2
+ 5x + 2) ln x dx =
+
dx 3 2
3
3
x 5x2
x 5x2
= + + 2x ln x − +
3 2 3 2
+ 2x
ln x dx
1
+ 2x
x
dx = · · · · · ·
3. We use trigonometric substitution x = 2 sin t. Then dx = 2 cos t dt and
x
I ≡
2 2
4 sin t
√ dx = 2 cos t dt
4 − x2 2 cos t
= 4 sin 2
t dt = 2 (1 − cos 2t) dt = 2t − sin 2t + C
Notice that t = sin −1 (x/2) and
sin 2t = 2 sin t cos t = 1
2 (2 sin t) 4 − (2 sin t) 2 = 1
2 x 4 − x 2 .
−1 x
So the final answer is I = 2 sin
2
4. Let u = 1 + √ x. Then du = 1
2 √ dx and
x
Please finish this.
3 1 + √ x
√ x
x
− 4 − x2 + C.
2
√3 dx = u 2 du = 2
1
u 1/3 du = · · · · · ·

5. Use the partial fractions decomposition
Please finish this.
1
x 2 + 2x − 15 =
1
(x − 3)(x + 5)
1 1 1
= − .
8 x − 3 x + 5
6. Use the trigonometric substitution x = (3/2) tan t. Please continue and finish this.
7. Use the partial fractions decomposition of the form
1
4x 3 + x =
Please continue and finish this.
1
x(4x 2 + 1)
= A
x
+ Bx + C
4x 2 + 1 .
8. Use the trigonometric substitution x = cos t. Please continue and finish this.
9. Note that x 4 − x = x(x 3 − 1) = x(x − 1)(x 2 + x + 1). Use the partial fractions
decomposition of the form
2x 3 − x 2 − 2x − 2
x 4 − x
Please continue and finish this.
= A
x
B Cx + D
+ +
x − 1 x2 + x + 1 .
Determine which of the following improper integrals converge:
1.
2.
3.
4.
5.
1
−1
−1
−∞
∞
e
∞
−∞
1
0
dx
. Divergent.
(x + 1) 3/2
dx
. Convergent.
1 + x2 dx
. Divergent.
x ln x
xe −x2
dx. Convergent.
dx
x √ . Divergent. 6.
x
π/2
sec x dx. Divergent.
0
2

Sketch the curves described by parametric or polar equations and write down
the integrals representing their lengths (do not evaluate these integrals):
1. We have dx/dt = e t , dy/dt = −2e −2t . So the arc length is
1
−1
(dx/dt) 2 + (dy/dt) 2 dt =
1
2. dx/dt = 6 cos 3t, dy/dt = −6 sin 3t. So the arc length is
π/2
0
−1
6 2 cos 2 3t + 6 2 sin 2 3t dt
e 2t + 4e −4t dt.
4. We have x = r cos θ = cos 3θ cos θ and y = r sin θ = cos 3θ sin θ. So
dx/dθ = −3 sin 3θ cos θ − cos 3θ sin θ, dy/dθ = −3 sin 3θ sin θ + cos 3θ cos θ
So the required arc length is
L =
=
2π
0
2π
0
The required area is 1
2 C
(−3 sin 3θ cos θ − cos 3θ sin θ) 2 + (−3 sin 3θ sin θ + cos 3θ cos θ) 2 dt
9 sin 2 3θ + sin 2 2π
3θ dθ = 1 + 8 sin
0
2 3θ dθ.
r 2 dθ = 1
π/6
2 −π/6
Solve each of the following differential equations:
1. Separate variables and integrate:
e −y
dy =
2. Using the integratiing factor ex2, we have
d
dx (ex2y)
= (x + 1)e x2 +2x
cos 2 3θ dθ. Please compute this integral.
or e x2
y =
Let u = (x + 1) 2 ≡ x 2 + 2x + 1. The last integral becomes
Please finish this.
1
2
sin x dx. Please finish it.
e u−1 du = 1
2 eu−1 + C.
3
(x + 1)e x2 +2x dx.

3. This is a homogeneous equation. As usual, let u = y/x. Then y = xu, which gives
dy/dx = u + x(du/dx). The original equation becomes
u + x du 1 + u 1
= =
dx u + u2 u
or x du 1 − u2
= .
dx u
Separating variables, we have
u du dx
= . Please finish.
1 − u2 x
4. Answer: y = C1e (−2+√ 7)x + C2e (−2−√ 7)x . Notice that r = −2 ± √ 7 are roots of
r 2 + 4r − 3 = 0.
5. Answer: y = −x + C1e x + C2e −x .
6. Write M = e x sin y + 2x and N = e x cos y + 2y. We can check
∂M/∂y = ∂N/∂x = e x cos y.
Hence this equation is exact. We look for a function u = u(x, y) satisfying
∂u
∂x = M ≡ ex sin y + 2x,
From the first equation we have
u =
∂u
∂y = N ≡ ex cos y + 2y.
(e x sin y + 2x) dx = e x sin y + 2x 2 + h(y).
Substitutuing this in the second equation, we obtain h ′ (y) = 2y, from which we get
h = y 2 + C0. So the final answer (in the form u = C) is e x sin y + 2x 2 + y 2 = C.
7. The general solution to the homogeneous equation y ′′ + y ′ − 2y = 0 is yH =
C1e −2x + C2e x . The final answer is of the form y = yH + yS. To find a special
solution yS, try yS = A cos(2x) + B sin(2x). Please complete these steps.
8. The final answer is y = yH + yS, where yH is the same as the one in Question 7
above. To find a particulra solution yS, try yS = Ae −x . Please complete this.
9. The final answer is y = yH + yS. where yH is the same as the one in Question 7
above. To find a particulra solution yS, try yS = Ae 3x . Please complete this.
Determine whether the given sequence is convergent or divergent:
1.
(−1) n e n
n 2
. The sequence diverges to infinity.
4

2.
3.
n 2n
. The sequence diverges to infinity in view of
1 + n
2 n
n n
. The sequence converges to zero.
2n
1 + n
→ 2 as n → ∞.
Determine whether the given series is conditionally convergent, absolutely convergent,
or divergent:
1. ∞
n=1
2. ∞
n=1
3. ∞
n=1
1 ∞
≤
nen n=1
4n − 1
n 3
1
e n
≤ ∞
n=1
n + 2n ∞
≤
1 + 3n n=1
= 1
e − 1
4n ∞
= 4
n3 n=1
2 n + 2 n
3 n
4. The series converges. Actually, ∞
5. This series is absolutely convergent.
6. Notice that
n n n + 1
=
2n + 1
n + 1
< ∞. So the series converges.
1
< ∞. So the series converges.
n2 (= 4) < ∞. So the series converges.
n=1
3 2n+1
n!
2n + 1
= 3 ∞
→ 1
2
n=0
9 n
n! = 3e9 .
as n → ∞.
Since 1/2 < 1, the root test tells us the convergence of the given series.
7. Note that cos nπ is nothing but (−1) n . So
∞
n=1
n cos nπ ∞
=
1 + n2 n=1 (−1)n n
1 + n2 which is convergent in view of the alternating series test. This series is not absolutely
convergent. Indeed
∞ n
= ∞.
n=1 1 + n2 8. Note that
n
n2n (1 + 2n2 =
) n n2
Hence the series is convergent.
1
→
1 + 2n2 2
5
as n → ∞.

9. Notice that, for large n,
So
∞
n=1
sin π
n
≈ π
n
sin x
because lim
x→0 x
sin π
, comparable to the divergent series
n
10. Use the integral test and compute
∞
11. The series is conditionally convergent.
2
dx
x ln x
12. Since n/ ln n → ∞ as n → ∞, the series diverges.
∞
n=1
= 1
.
1
, diverges as well.
n
= · · · = ∞. The series diverges.
In each part, find all x for which the given series is convergent:
1.
2.
3.
4.
5.
∞ (x − 1) n
n=1
∞
n=1
∞
n=1
n!
n 3 (x − 5) n
x n
5 n n 5
∞ (4x + 1) n
n=1
∞
n=1
n 2
n(x − 4) n
2n 2 + 3
Answer: (−∞, ∞)
Answer: (4, 6)
Answer: [−5, 5].
Answer: [−1/2, 0].
Answer: [−3, 5).
Find the Taylor series of f(x) at x = a and its interval of convergence:
Let’s recall a list of well-known series expansions that one should learn by heart:
1
1 − x =
∞
x n
n=0
ln
1
1 − x =
6
∞
n=1
x n
n
e x =
∞
n=0
x n
n!

sin x =
∞
(−1)
n=0
n x2n+1
(2n + 1)!
cos x =
If possible, we make use of any one of these identities.
∞
(−1)
n=0
n x2n
(2n)! .
1. In the following manipulation, keep in mind that we are trying to use the second
identity above
ln x = ln(3 − (3 − x)) = ln 3(1 − (1 − x/3)) = ln 3 + ln(1 − (1 − x/3))
= ln 3 − ∞
n=1
(1 − x/3) n
(Note that 1 − x/3 = ((−1)/3) (x − 3).)
2. With a = −1, we have
e 2x+3 = e 2(x+1)+1 = e e 2(x+1) = e ∞
3. Again we try to ue the second identity
4. At x = π/3, we have
5. This is easy:
n
= ln 3 − ∞
n=0
n=1
2 n (x + 1) n
ln(2 + x 2 ) = ln 2(1 − (−x 2 /2)) = ln 2 − ln
= ln 2 − ∞
n=0
n=1
(−x/2) n
n
n!
= ln 2 − ∞
(−1) n
3 n n (x − 3)n .
= e ∞
n=0
1
1 − (−x/2)
n=1
(−1) n
2 n n xn .
2 n
n! (x + 1)n .
π
sin x = sin
3 +
x − π
= sin
3
π
3 cos
x − π
+ cos
3
π
3 sin
x − π
3
√ ∞ 3 (−1)
=
2
n
x −
(2n)!
π
2n + =
3
1
∞ (−1)
2
n
x −
(2n + 1)!
π
2n+1 .
3
xe x = x ∞
n=0
6. We use the double angle formula
x n
n! =
∞
n=0
sin x cos x = 1 1 ∞
sin(2x) =
2 2 n=0
= 1 ∞
2 n=0
n=0
x n+1
n! =
∞
n=1
(−1) n (2x) 2n+1
(2n + 1)!
(−1) n 2 2n+1
(2n + 1)! x2n+1 = ∞
7
n=0
x n
(n − 1)! .
(−1) n 2 2n
(2n + 1)! x2n+1 .

Find the binomial series of the given function
We use the binomial expansion formula (1 + x) α = ∞
1.
1
(1 − x) 3 = (1 + (−x))−3 = ∞
So we have
−3
=
n
(−3)(−3 − 1)(−3 − 2) · · · (−3 − n + 1)
n=0
n=0
−3
(−x)
n
n with
n!
n 3 × 4 × 5 × · · · × (n + 2)
= (−1)
1 × 2 × 3 × · · · × n
1
= (1 + (−x))−3
(1 − x) 3
∞
n (n + 1)(n + 2)
= x (−1) (−x)
2
n =
n=0
2. x √ 1 − x = x(1 + (−x)) 1/2 = x ∞
1/2
(−1)
n
n =
n=1
1/2
n
α
x
n
n to each part:
n (n + 1)(n + 2)
= (−1)
1 × 2
∞
n=0
(−1) n x n . For n ≥ 1,
1/2(1/2 − 1)(1/2 − 2) · · · (1/2 − n + 1)
(−1)
n!
n
= (−1)n−1
2 n
(n + 1)(n + 2)
x
2
n+1 .
1 × 3 × 5 × · · · × 2n − 1
(−1)
n!
n (2n − 1)!!
= −
2n .
n!
8