MATH 31B: MIDTERM 2 REVIEW
JOE HUGHES
1. Trigonometric Integrals
1. Evaluate
R
sin2 x dx and
R
cos2 x dx.
Solution: Recall the identities
cos2 x =
1 + cos(2x)
2
sin2 x =
1 − cos(2x)
2
Using these formulas gives
Z
Z
x sin(2x)
1 − cos(2x)
sin2 x dx =
dx = −
+C =
2
2
4
and similarly
Z
Z
1 + cos(2x)
x sin(2x)
cos2 x dx =
dx = +
+C =
2
2
4
x sin x cos x
−
+C
2
2
x sin x cos x
+
+C
2
2
The last step of both integrals used the formula sin(2x) = 2 sin x cos x.
R
2. Evaluate sin2 x cos3 x dx.
Solution: The key idea for evaluating trig integrals with odd powers is to use the identity
sin2 x + cos2 x = 1 to try to set up a substitution. In this case, we get
Z
Z
Z
sin2 x cos3 x dx = sin2 x cos2 x cos x dx = sin2 x(1 − sin2 x) cos x dx
Now the integral is written in a way that makes a u-substitution possible. Let u = sin x,
so du = cos x dx. Then the integral becomes
Z
Z
u3 u5
sin3 x sin5 x
2
2
u (1 − u ) dx = u2 − u4 du =
−
+C =
−
+C
3
5
3
5
One of the challenges of trig integrals is that the answer can take several different forms,
depending on how many trig identities one chooses to use. So as a sanity check, let’s take
the derivative of our answer:
i
d h sin3 x sin5 x
−
+ C = sin2 x cos x − sin4 x cos x = sin2 x cos x − sin2 x sin2 x cos x
dx
3
5
1
2
JOE HUGHES
= sin2 x cos x − sin2 x(1 − cos2 x) cos x = sin2 x cos x − sin2 x cos x + sin2 x cos3 x
= sin2 x cos3 x
which is what we started with.
R
3. Evaluate cos4 x dx.
Solution: To compute this integral, we can repeatedly use the identity from problem 1.
Thus
1 + cos(2x) 2 1 h
i
= 1 + 2 cos(2x) + cos2 (2x)
cos4 x = (cos2 x)2 =
2
4
1 + cos(4x) i
1h
= 1 + 2 cos(2x) +
4
2
and
Z
Z
1
1 + cos(4x)
cos4 x dx =
1 + 2 cos(2x) +
dx
4
2
Z
Z
Z
1
1
3
dx +
2 cos(2x) dx +
cos(4x) dx
=
8
4
8
3
1
1
= x + sin(2x) +
sin(4x) + C
8
4
32
2. Trig Substitutions
1. Evaluate
R
2
√x
9−x2
dx.
Solution: Make the substitution x = 3 sin θ, so that dx = 3 cos θ dθ. Then
Z
Z
Z
9 sin2 θ
x2
√
dx =
3 cos θ dθ = 9 sin2 θ dθ
3 cos θ
9 − x2
9
= [θ − sin θ cos θ] + C
2
using the result from the previous section. Now θ = sin−1 x3 , sin θ = x3 , and
√
cos θ =
Z
9−x2
,
3
so
√
x2
9 h −1 x x 9 − x2 i
9
x 1 p
√
dx =
sin
−
+ C = sin−1 − x 9 − x2 + C
2
3
3
3
2
3 2
9 − x2
R
dx
2. Evaluate √12x−x
.
2
Solution: At first glance, it doesn’t look like we can do a trig substitution. But observe
that
(x − 6)2 = x2 − 12x + 36
MATH 31B: MIDTERM 2 REVIEW
3
hence
12x − x2 = 36 − (x − 6)2
Therefore
Z
dx
√
=
12x − x2
Z
Z
dx
p
=
36 − (x − 6)2
√
du
36 − u2
by taking u = x − 6.
Now we can do a trig substitution: let u = 6 sin θ. Then du = 6 cos θ dθ, hence
Z
Z
du
6 cos θ
u
x−6
√
=
dθ = θ + C = sin−1 + C = sin−1
+C
2
6 cos θ
6
6
36 − u
3. Partial Fractions
1. Evaluate
R
x−1
(x−2)2 (x−3)
dx.
Solution: The numerator has degree less than the denominator, so we can use partial
fractions. Write
A1
A2
B
x−1
=
+
+
2
2
(x − 2) (x − 3)
x − 2 (x − 2)
x−3
After clearing denominators, we get
x − 1 = A1 (x − 2)(x − 3) + A2 (x − 3) + B(x − 2)2
Now plug in x = 2 to obtain 1 = −A2 , so A2 = −1. Similarly, if we plug in x = 3 then we
get 2 = B.
To determine A1 , let’s look at the coefficient of x2 on both sides. On the left-hand side,
this coefficient is zero, while on the right-hand side the coefficient is A1 + B. Therefore
A1 = −B = −2. So our integral becomes
Z
Z
Z
Z
x−1
dx
dx
dx
dx = −2
−
+2
(x − 2)2 (x − 3)
x−2
(x − 2)2
x−3
= 2 ln |x − 3| − 2 ln |x − 2| +
2. Evaluate
R
5x
(x+2)(x2 +1)
1
+C
x−2
dx.
Solution: When we do partial fractions with an irreducible quadratic factor like x2 + 1, the
corresponding term in the expansion should have a linear polynomial in the numerator. So
write
5x
A
Bx + C
=
+ 2
(x + 2)(x2 + 1)
x+2
x +1
Now clear the denominators to get
5x = A(x2 + 1) + (Bx + C)(x + 2)
4
JOE HUGHES
Next, plug in x = −2:
5(−2) = A(4 + 1)
or A = −2.
To find B, look at the coefficient of x2 . On the left-hand side this coefficient is zero, while
on the right hand side it is A + B. Therefore B = −A = 2. Finally, to find C we look at
the constant coefficient on both sides to get 0 = A + 2C, so C = − A2 = 1.
Therefore
Z
Z
Z
Z
Z
Z
dx
2x + 1
dx
2x
dx
5x
dx = −2
+
dx = −2
+
dx+
2
2
2
2
(x + 2)(x + 1)
x+2
x +1
x+2
x +1
x +1
= −2 ln |x + 2| + ln(x2 + 1) + tan−1 (x) + C
3. Evaluate
sec2 θ
tan2 θ−1
R
dθ.
Solution: One might be tempted to try to use the trig identity sec2 θ = 1 + tan2 θ, but I
don’t think that helps here. So instead let’s start with a substitution: let u = tan θ, so
that du = sec2 θ dθ. Then
Z
Z
Z
du
du
sec2 θ
dθ =
=
2
2
u −1
(u − 1)(u + 1)
tan θ − 1
Now we can proceed using partial fractions. Write
1
A
B
=
+
(u − 1)(u + 1)
u−1 u+1
and clear denominators to obtain
1 = A(u + 1) + B(u − 1)
Plugging in u = 1 gives 1 = 2A, so A = 12 . Similarly, plugging in u = −1 gives B = − 12 .
Therefore
Z
Z
Z
du
1
du
1
du
1
1
=
−
= ln |u − 1| − ln |u + 1| + C
(u − 1)(u + 1)
2
u−1 2
u+1
2
2
and plugging in u = tan θ gives
Z
sec2 θ
1
1
dθ = ln | tan θ − 1| − ln | tan θ + 1| + C
2
2
tan2 θ − 1
4. Improper Integrals
1. Determine whether
R1
0
7
x− 8 dx converges, and if so, evaluate it:
7
Solution: The function x− 8 has an infinite discontinuity at x = 0, so this is an improper
integral. We can evaluate it using the limit definition:
Z 1
Z 1
h 1 i1
7
7
1
x− 8 dx = lim
x− 8 dx = lim 8x 8 = lim 8 − 8a 8 = 8
0
a→0+
a
a→0+
a
a→0+
MATH 31B: MIDTERM 2 REVIEW
5
Therefore the integral converges, and the value of the integral is 8.
Z ∞
dx
converges.
2. Determine whether
1
1
2x 2 + x
1
1
Solution: 2x 2 ≤ x for x ≥ 4, hence 2x 2 + x ≤ 2x and
1
1
≥
2x 2 +x
1
2x
for x ≥ 4.
We can break our integral up into two pieces:
Z ∞
Z 4
Z ∞
dx
dx
dx
=
+
1
1
1
1
1 2x 2 + x
4
2x 2 + x
2x 2 + x
The first integral is just some finite number, so it doesn’t affect the convergence or divergence of the original integral. For the second integral, the comparison test gives
Z ∞
Z ∞
h
iR 1
dx
1
dx
=
lim ln(x) =
lim ln(R) − ln(4) = ∞
≥
1
2x
2 R→∞
2 R→∞
4
4
4
2x 2 + x
Therefore
∞
Z
2x + x
4
diverges, which shows that
dx
1
2
∞
Z
1
dx
1
2
2x + x
also diverges.
5. Numerical Integration
1. Find N such that
Z
1
2
− 12
cos x dx − TN ≤ 10−5
Solution: The formula for the error bound for the trapezoid rule is given by
Error(TN ) ≤
K2 (b − a)3
12N 2
where K2 = max |f 00 (x)|.
x∈[a,b]
If f (x) = cos x, then f 0 (x) = − sin x and f 00 (x) = − cos x. Therefore
K2 =
max | cos x| = 1
x∈[− 12 , 12 ]
hence
Error(TN ) ≤
( 21 − (− 21 ))3
1
=
2
12N
12N 2
6
JOE HUGHES
We want the error to be at most 10−5 . Thus
1
≤ 10−5
12N 2
and solving for N gives
r
N≥
105
≈ 91.2
12
so any N ≥ 92 will do.
How could we handle the last part without using a calculator? Using the fact that
we see that any N ≥
√
105
105
≤
= 104
12
10
104 = 102 = 100 suffices.
6. Arc length
For t ∈ [0, 12 ], let f (t) be the arclength of the curve
y=
x1−t
xt+1
−
2(1 − t) 2(t + 1)
on [0, 1]. Find the maximum of f (t) on [0, 21 ] and the point at which it is attained.
Solution: Recall that the arclength formula is
Z 1r
dy 2
1+
dx
dx
0
By the power rule,
dy
x−t xt
=
−
dx
2
2
hence
1+
dy 2
dx
=1+
x−2t 1 x2t
x−2t 1 x2t x−t xt 2
− +
=
+ +
=
+
4
2
4
4
2
4
2
2
Plugging this into the arc length formula gives
Z
f (t) =
0
1
xt x−t
1
1
1
+
dx =
+
=
2
2
2(1 + t) 2(1 − t)
1 − t2
Therefore f is increasing on [0, 12 ], so has a maximum of
4
3
at t = 12 .
MATH 31B: MIDTERM 2 REVIEW
7
7. Fluid Force
1. Find the fluid force on a metal plate bounded by the parabola y = −x2 and the line
y = −1. The surface of the liquid is the line y = 0, and the liquid has pressure ρ.
Solution: The formula for fluid force is
1
Z
F = ρg
yf (y) dy
0
√
where f (y) is the width at depth y. In this case, f (y) = 2 y. Therefore
Z 1
Z 1
h 2 5 i1 4
3
√
y y dy = 2ρg
y 2 dy = 2ρg y 2 = ρg
F = 2ρg
5
5
0
0
0
8. Taylor Polynomials
1. Let f (x) = e−x . Find the third Taylor polynomial T3 (x) for f centered at x = 0, and
use the error bound to find the maximum value for |f (0.1) − T3 (0.1)|.
Solution: Recall that
T3 (x) = f (0) + f 0 (0)x +
f 00 (0) 2 f (3) (0) 3
x +
x
2
6
Therefore we need to take 3 derivatives:
f 0 (x) = −e−x
f 00 (x) = e−x
f (3) (x) = −e−x
Plugging in x = 0, we get
T3 (x) = 1 − x +
x2 x3
−
2
6
Next, recall that the error bound formula is
K
K|0.1 − 0|4
=
4!
24 · 104
where K is an upper bound for |f (4) (x)| on [0, 0.1]. To get the best possible error bound,
we take
K = max |f (4) (x)|
|f (0.1) − T3 (0.1)| ≤
x∈[0,0.1]
Now
f (4) (x)
=
e−x ,
which is a decreasing function. Therefore
K = max |f (4) (x)| = e0 = 1
x∈[0,0.1]
hence
|f (0.1) − T3 (0.1)| ≤
1
≈ 4.16 × 10−6
24 · 104
On the midterm, I think that it would suffice to leave the answer as
1
.
24·104