c = ln ∣2 cos θ + sin θ

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1041微微微甲甲甲01-04班班班期期期末末末考考考解解解答答答和和和評評評分分分標標標準準準 1. (16%) Compute the following integrals.

(a) ∫

cos θ − 2 sin θ

2 cos θ + sin θdθ. (b) ∫

√ 1 + x²

x² dx.

Solution:

1. (Method I)Let u = 2 cos θ + sin θ ⇒ du = cos θ − 2 sin θdθ

⇒ ∫

cos θ − 2 sin θ 2 cos θ + sin θdθ =∫

1

udu = ln ∣u∣ + c = ln ∣2 cos θ + sin θ∣ + c (Method II) Let t = tanθ

2 ⇒dθ = 2

1 + t²dt, sin θ = 2t

1 + t², cos θ =1 − t² 1 + t²

⇒ ∫

(¹^−t

2

1+t²) −2 (₁^2t_+t₂) 2 (¹₁^−t_+t²₂) + (₁^2t_+t₂)

⋅ 2

1 + t²dt =∫

2 − 2t²−8t (2 + 2t − 2t²)(1 + t²)

dt 1 − t²−4t

(1 + t − t²)(1 + t²)

= At + B

1 + t² +

Ct + D 1 + t − t² =

(At + B)(1 + t − t²) + (Ct + D)(1 + t²) (1 + t − t²)(1 + t²)

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

C − A = 0 D − B + A = −1 A + B + C = −4 B + D = 1

⇒

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

A = −2 B = 0 C = −2 D = 1

⇒ ∫

2 − 2t²−8t (2 + 2t − 2t²)(1 + t²)

dt =∫ − 2t 1 + t²−

2t − 1 1 + t − t²dt

= −ln ∣t²+1∣ + ln ∣−t²+t + 1∣ + c = ln ∣1 − t²+t

1 + t² ∣ +c = ln ∣1 − t² 1 + t² +

1 2(

2t 1 + t²)∣ +c

=ln ∣cos θ +1

2sin θ∣ + c = ln ∣2 cos θ + sin θ∣ − ln 2 + c = ln ∣2 cos θ + sin θ∣ + c^′ (Method III)∫

sin(α − θ)

cos(α − θ)dθ = −∫ tan(α − θ)d(α − θ), where tan α = 1 2

⇒ − ∫ tan(α − θ)d(α − θ) = − ln ∣sec(α − θ)∣ + c = ln ∣2 cos θ + sin θ∣ + c^′ 2. Let x = tan θ ⇒ dx = sec²θdθ

∫

√ 1 + x²

x² dx =∫ sec³θ tan²θdθ =∫

1 sin²θ cos θdθ (Method I) ∫

1

sin²θ cos θdθ =∫

cos θ sin²θ cos²θdθ Let u = sin θ ⇒ du = cos θdθ ⇒ u = x

√ 1 + x²

⇒ ∫

cos θ

sin²θ cos²θdθ =∫ 1 u²(1 − u²)

du =∫ 1 u²+

1 1 − u²du

= ∫ 1 u²+

1 2(

1 1 − u+

1

1 + u)du = −1

u+ln ∣1 + u 1 − u∣ +c

= −

√ 1 + x²

x +

1 2ln ∣

√

1 + x²+x

√

1 + x²−x∣ +c

= −

√ 1 + x²

x +

1 2ln

RR RR RR RR RR RR RR R (

√

1 + x²+x)² (1 + x²) −x²

RR RR RR RR RR RR RR R

+c = ln ∣

√

1 + x²+x∣ −

√ 1 + x²

x +c

(Method II) ∫ 1

sin²θ cos θdθ =∫

sin²θ + cos²θ sin²θ cos θ dθ =∫

1 cos θ +

cos θ sin²θdθ

= ∫ sec θ + cot θ csc θdθ = ln ∣ sec θ + tan θ∣ − csc θ + c

=ln ∣x +

√

1 + x²∣ −

√ 1 + x²

x +c (Method III) ∫

1

sin²θ cos θdθ =∫ csc²θ sec θdθ = −∫ sec θd cot θ = − sec θ cot θ +∫ sec θ tan θ cot θ

= −csc θ + ln ∣sec θ + tan θ∣ + c = ln ∣x +

√

1 + x²∣ −

√ 1 + x²

x +c

Page 1 of 10

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(Method IV) Let x = sinh θ ⇒ dx = cosh θdθ

∫

√ 1 + x²

x² dx =∫ coth²θdθ =∫ csch²θ + 1dθ = − coth θ + θ + c =

√ 1 + x²

x +sinh⁻¹x + c (∵sinh⁻¹x = ln ∣x +

√ 1 + x²∣)

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2. (16%) Compute the following integrals.

(a) ∫

√

1 − x²sin⁻¹xdx. (b) ∫ ln(√

x +√

1 + x)dx.

Solution:

note ∶

∫ u dv = uv −∫ v du

(a )

(i) Let x = sin θ, dx = cos θdθ

∫

√

1 − x²sin⁻¹x dx

= ∫ (cos²θ)θ dθ =∫ (

1 + cos 2θ 2 )θ dθ

= ∫ 1

2θ dθ +∫ 1

2θ cos (2θ) dθ Let u = θ; v = 1

2sin 2θ. So we can use integration by part to get

∫ 1

2θ dθ +∫ 1

2θ cos (2θ) dθ = 1 4θ²+

1 2[

1

2θ sin (2θ) −∫ 1

2sin (2θ)dθ]

= 1 4θ²+

1

4θ sin (2θ) +1

8cos (2θ) + C =1

4(sin⁻¹x)²+ 1

4(sin⁻¹x) sin (2 sin⁻¹x) +1

8cos (2 sin⁻¹x) + C, C ∈ R

= 1

4(sin⁻¹x) +1

2(sin⁻¹x)x

√ 1 − x²−

1

4x²+C^′, C^′=C +1 8 (ii)∫ cos²θ dθ = 1

2θ +1

4sin (2θ).

Let u = θ; v = 1 2θ +1

4sin (2θ). So we can get

∫ θ cos²θ dθ = (1 2θ + 1

4sin (2θ))θ −∫ 1 2θ +1

4sin (2θ) dθ =1 4θ²+

1

4θ sin (2θ) +1

8cos (2θ) + C, C ∈ R (iii) Let x = sin θ. So we can get

∫

√

1 − x²dx =∫ cos²θ dθ = 1 2θ +1

4sin (2θ) =1

2(sin⁻¹x) +1 2x

√ 1 − x². Let u = sin⁻¹x; v =1

2(sin⁻¹x) +1 2x

√ 1 − x². Using integration by part,we can get

∫

√

1 − x²sin⁻¹x dx = [1

2(sin⁻¹x +1 2x√

1 − x²)](sin⁻¹x) −∫ ( 1

2(sin⁻¹x) +1 2x√

1 − x²) 1

√

1 − x² dx

= 1

2(sin⁻¹x)²+ 1 2x

√

1 − x²sin⁻¹x −1

4(sin⁻¹x)²− 1

4x²+C =1

4(sin⁻¹x)²+ 1

2(sin⁻¹x)x

√ 1 − x²−

1

4x²+C, C ∈ R (b )

Let x = tan²θ, dx = 2 tan θ sec²θdθ

∫ log (√ x +

√

1 + x) dx

= ∫ log (tan θ + sec x)2 tan θ sec²θ dθ =∫ log (tan θ + sec θ) d(sec²θ)

=sec²θ log (tan θ + sec θ) −∫ sec³θ dθ = sec²θ log (tan θ + sec θ) −sec θ tan θ + log (sec θ + tan θ)

2 +C

=(sec²θ −1

2)log (sec θ + tan θ) −1

2sec θ tan θ + C = (tan²θ +1

2)log (sec θ + tan θ) −1

2sec θ tan θ + C

=(x +1 2)log(√

x +√

1 + x) −1 2

√x√

1 + x + C, C ∈ R

Page 3 of 10

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3. (8%) Suppose that f (x) is a polynomial whose coefficients are integers, and

∫

∞ 0

f (x)

(x + 1)²(4x²+1)dx = 2 ln 2 + 1.

Find f (x).

Solution:

The degree of f (x) must be less or equal than three.

However, if the degree of f (x) equal three, and the value after the integral is divergent. Consequently, the degree of f (x) = 2 by estimation.

By using partial fractions method, f (x)

(x + 1)²(4x²+1)= A (x + 1)²+

B x + 1+

Cx + D

4x²+1. (3 points) We integrate both sides,

2 ln 2 + 1 = lim

p→∞∫

p 0

A

(x + 1)²dx + lim

p→∞∫

p 0

B

x + 1dx + lim

p→∞∫

p 0

Cx + D 4x²+1dx.

= lim

p→∞[

−A

x + 1 + B ln ∣ x + 1 ∣ + C

8 ln ∣ 4x²+1 ∣ + D

2 tan⁻¹(2x)]^p₀ (1 points)

By using the limits must be exist, i.e. the natural log term must be equal to 2 ln 2 , and we found C = −4B.

By campring the coefficents with (2 ln 2 + 1), We found that A = 1, B = −2, C = 8 and D = 0.

(Because of A, B, C and D are all integers in the problem, the values of A, B, C and D which we found are correct.) Therefore,

f (x) = [ 1 (x + 1)²+

−2 x + 1+

8x

4x²+1] × [(x + 1)²(4x²+1)].

=12x²+6x − 1. (4 points)

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4. (16%) Let M_n=

n

∑

i=1

1 n +

√

n(i −¹₂) .

(a) Recognize Mn as the Midpoint approximation of a definite integral and compute I = lim

n→∞Mn.

(b) Write down the Trapezoidal approximation of the integral, T_n, and the right point approximation, R_n. (c) For any value of n, list Mn, Tn, Rn, and I in increasing order.

Solution:

(a) I = lim

n→∞M_n= lim

n→∞

n

∑

i=1

1 1 +

√

i−¹2

n

1 n

= ∫

1 0

1 1 +√

xdx (2pt) ( Let√

x = u, dx = 2udu )

= ∫

1 0

2u

1 + udu =∫

1 0

2 − 2

1 + udu = (2u − 2 ln ∣1 + u∣)∣

1 0

=2 − 2 ln 2 (2pt) (b)

T_n= 1 2n

n

∑

i=1

( 1 1 +

√

i n

+ 1 1 +

√

i−1 n

) (4pt)

Rn= 1 n

n

∑

i=1

( 1 1 +

√

i n

) (4pt)

(c)

Rn<Mn<I < Tn (4pt)

Page 5 of 10

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5. (8%) A sphere of radius 1 overlaps a smaller shpere of radius r (0 < r < 1) in such a way that their intersection is a circle of radius r, i.e. a great circle of the small sphere. Find r so that the volume inside the small sphere and outside the large sphere is as large as possible.

Solution:

Let C1 be a circle with center (0, 0) and radius 1, and let C2 be a circle with center (

√

1 − r², 0) and radius r.

Then the desired volume V (r) is equal to the volume of the solid obtained by rotating the region inside C2

and outside C1 about the x-axis, so V (r) = 1

2⋅ 4 3πr³− ∫

1

√1−r²π(

√ 1 − x²)

2

dx (4pts)

Applying the Fundamental Theorem of Calculus, we can obtain dV

dr = 2πr²+π⎛

⎝ 1 − (√

1 − r²)

2⎞

⎠

⋅ d dr

√ 1 − r²

= πr²⎛

⎝

2 − r

√ 1 − r²

⎞

⎠

(2pts)

The derivative dV

dr is 0 when 2 = r

√

1 − r², that is, r = 2

√5. (1pt)

Since

rlim→0⁺V (r) = 0, lim

r→1⁻V (r) = 0, V ( 2

√5) =2π( 1

√5− 1 3) >0

we have the maximum value V (r) = 2π( 1

√5− 1

3)when r = 2

√5. (1pt)

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6. (16%) Let a and b be fixed numbers.

(a) Find parametric equations for the curve C that consists of all possible positions of the point P in the figure using the angle θ as the parameter.

(b) What are the horizontal asymptotes of the curve C?

(c) Let R be the region bounded by C and its horizontal asymptotes. Find the area of R if it is finite..

(d) The region is R rotated about the y-axis. Find the volume of the resulting solid if it is finite.

Solution:

(a) Q = (b cos θ, b sin θ) (1分) B = (a sec θ, 0) (1分)

→P = (a sec θ, b sin θ) (2分) θ ∈ [0, 2π]不包含{π

2,3π

2 } (少寫不扣分) (b) 作法一：Use. (a

x)²+ ( y

b)²=1 (2分) and x → ±∞, y = ±b (1分) 作法二：y = ±b

√ 1 − (a

x)², ∣x∣ ≥ (2分) ⇒ lim

x→∞y = ±b (1分) 作法三：P 的x座標a sec θ → ±∞ ⇔ θ →π

2 or 3π

2 (2分), then y座標 b sin θ → ±b (1分) 作法四：作圖 → y = ±b (3分)

(c) (列式2分；計算3分) R = 4{ab +∫

∞ a

b − b

√ 1 − (a

x)²dx} (列式2分)

∫

√ 1 − (a

x)²dx =∫

√ x²−a²

x² dx^u⁼

√x²−a²

= ∫

u² u²+a²du =

√

x²−a²−a tan⁻¹(

√

x²−a²a)

=4{ab + b(x −

√

x²−a²+a tan⁻¹(

√ x²−a²

a ))^∞_a } =2πab OR

R = 4[ab +∫

∞

a (b − y)dx] = 2πab (列式2分)

∫

π/2 0

(b − b sin θ)a tan θ sec θdθ = ab∫

π/2 0

(tan θ sec θ − θ²θ)dθ

=ab(sec θ − tan θ + θ)^π₉^/2=ab(π 2 −1) (d) (計算2分；答案2分)

x = a

√

1 − (^y_b)²

⇒ Volume = 2∫

b 0

πx²dy = 2πa²b²∫

b 0

dy

b²−y² diverges OR

Volume=2π

π 2

0a²sec²θb cos θdθ = 2a²b∫

π 2

0

sec θdθ = 2a²b ln ∣ sec θ + tan θ∣

π 2

0 diverges

Page 7 of 10

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7. (16%)

(a) Sketch the curve r = 1 + 2 cos θ.

(b) Compute the area of the region that is inside the larger loop of the curve r = 1 + 2 cos θ and outside the smaller loop of the curve r = 1 + 2 cos θ.

(c) Let C be the smaller loop of r = 1 + 2 cos θ, and rotate C about the x-axis. Find the area of the resulting surface.

Solution:

(a) 0 = 1 + 2 cos θ ⇒ cos θ =−1

2 ⇒θ = 2π 3 ,4π

3 Figure: (2 points)

The figure for r = 1 + 2 cos θ.

(b)A = A_total−Ainner circle

=2[1 2∫

2π 3

0

(1 + 2 cos θ)²dθ − 1 2∫

π

2π 3

(1 + 2 cos θ)²dθ] (4 points)

= ∫

2π 3

0

(1 + 4 cos θ + 4 cos²θ)dθ − ∫

π

2π 3

(1 + 4 cos θ + 4 cos²θ)dθ

= (3θ + 4 sin θ + sin 2θ) ∣

2π 3

0 − (3θ + 4 sin θ + sin 2θ) ∣^π2π 3

= (2π +3√ 3

2 ) − (π −3√ 3 2 )

=π + 3√

3. (3points) (c) A =∫ 2πyds

=2π∫

b a

r sin θ

√

r²+r^′2 dθ (2 points)

There are two ranges could describe this figure, i.e. from π to 4π

3 or from 2π 3 to π.

However, if we choose from 2π

3 to π, and the area would be negative because of r sin θ ≤ 0. Therefore, we choose the range which is from π to 4π

3 . A = 2π∫

4π 3

π

(1 + 2 cos θ) sin θ

√

(1 + 2 cos θ)²+ (2 sin θ)²dθ

=2π∫

4π 3

π

(1 + 2 cos θ) sin θ√

5 + 4 cos θ dθ Let u = cos θ, and du = − sin θ dθ.

θ from π to 4π

3 ⇒ u from -1 to -0.5.

=2π∫

−0.5

−1 (1 + 2u)√

5 + 4u (−du). (2 points) Then, Let t = 5 + 4u, and dt = 4du

u from -1 to -0.5 ⇒ t from 1 to 3.

=2π∫

3 1

(1 + 2 ⋅t − 5 4 )

√ t (−1

4)dt.

π(3√

3 − 2). (3 points)

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8. (16%) A candle is located at the origin O, a bug, P , crawls on the plane so that the angle between its velocity and the vectorÐ→

P O is always π 6.

(a) Suppose that the bug crawls at a curve with polar equation r = f (θ). Derive the differential equation that f (θ) satisfies.

(b) If the Cartesian coordinates for the bug’s initial position are (1, 0), solve for the curve r = f (θ).

(c) Compute the arc length function s(θ) and find lim

θ→∞s(θ).

Solution:

(a) (8pts)

Let φ be the angle between the tangent line at P and the x-axis.

Then

±tanπ

6 = tan(φ − θ) = tan φ − tan θ 1 + tan φ tan θ =

dy dx−tan θ 1 +^dy_dx⋅tan θ

=

r^′sin θ+r cos θ r^′cos θ−r sin θ−_{cos θ}^{sin θ} 1 + ^r_r^′′^{sin θ}cos θ^{+r cos θ}−r sin θ⋅_{cos θ}^{sin θ}

=

(r^′sin θ + r cos θ) cos θ − (r^′cos θ − r sin θ) sin θ (r^′cos θ − r sin θ) cos θ + (r^′sin θ + r cos θ) sin θ =

r r^′

⇒ { f^′(θ) =√ 3f (θ) f^′(θ) = −

√ 3f (θ)

(b) Case 1. If f^′(θ) =√

3f (θ), then

∫ 1

fdf = ∫

√ 3θdθ

⇒ln ∣f ∣ =

√ 3θ + C and so

f (θ) = C^′e^√^3θ (2pts) Since y(1) = 0 ⇒ f (0) = 1, we have

1 = f (0) = C^′e^√³^⋅0=C^′ Therefore, the solution is

f (θ) = e

√3θ

(1pt)

Case 2. If f^′(θ) = −

√

3f (θ), then, similarly, the solution is f (θ) = e⁻^√^3θ (c) Case 1. If f (θ) = e

√3θ

, then

s(θ) = ∫

θ 0

√ (f (t))

2

+ (d dtf (t))

2

dt (2pts)

= ∫

θ 0

√ (e^√^3t)

2

+ ( d dte^√^3t)

2

dt

= 2

√3(e^√^3θ−1) (2pts) Therefore,

lim

θ→∞s(θ) = ∞ (1pt) Case 2. If f (θ) = e⁻

√3θ

, then

s(θ) = 2

√

3(1 − e⁻

√3θ

), lim

θ→∞s(θ) = 2

√ 3

Page 9 of 10

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9. (8%) Solve the differential equation

x²y^′−y = 2x³e⁻¹^x, y(1) = 1.

Solution:

x²y^′−y = 2x³e⁻¹^x y^′−

1

x² =2xe⁻¹^x (2pt) I(x) = e^{∫ −}^x2¹^dx=e¹^x (1pt)

y(x) = 1

I(x)[∫ I (x)r(x)dx + c]

=e⁻^x¹[∫ e

1

x2xe⁻¹^xdx + c]

=e⁻^x¹[∫ 2xdx + c]

=e⁻^x¹(x²+c)(2pt) and y(1) = 1

⇒1 = e⁻¹(1 + c)

⇒c = e − 1 (1pt)

∴y = e⁻¹^x(x²+e − 1) (2pt)