(8%) Evaluate the integral Z 1 −1 ln  x

微甲01-05班 統一教學期末考解答

1. (8%) Evaluate the integral Z 1

−1

e^2x 1 + e^xdx.

Sol:

Let u = e^x, then du = e^xdx, and so Z 1

−1

e^2x

1 + e^xdx = Z e

e⁻¹

u 1 + udu

= Z e

e⁻¹

1 − 1 1 + udu

= [u − ln(1 + u)]^e_e−1

= e −1 e − 1.

2. (8%) Evaluate the integral Z 1

−1

ln

 x +√

1 + x²

 dx.

Sol:

Note that for −1 ≤ x ≤ 1,

ln(−x +p

1 + (−x)²) = ln 1 x +√

1 + x² = − ln(x +√

1 + x²).

So ln(x +√

1 + x²) is an odd function in [−1, 1], and thus the integral equals 0.

3. (8%) Evaluate the integral

Z dt

t −√

1 − t². Sol:

Set t = sin θ. Then the integral becomes to

Z cos θ

sin θ − cos θ dθ. Now Z cos θ

sin θ − cos θ dθ =

Z 1

tan θ − 1 dθ

Again we make a substitution. Let u = tan θ. du = sec²θdθ = (1 + u²)dθ. Hence

Z 1

tan θ − 1 dθ =

Z 1

(u − 1)(u²+ 1) du = Z 

1

2(u − 1) − u + 1 2(u²+ 1)

 du

Clearly,

Z 1

2(u − 1) du = 1

2ln |u − 1| + C₁

Z u + 1

2(u²+ 1) du = 1

4ln |u²+ 1| + 1

2arctan u + C₂ Hence one can get

Z 1

tan θ − 1 dθ = 1

2ln | tan θ − 1| − 1

4ln | tan²θ + 1| −1 2θ + C with C = C₁− C₂. Note

1

2ln | tan θ − 1| − 1

4ln | tan²θ + 1| = 1

2ln | tan θ − 1| − 1

2ln | sec θ| = 1

2ln | sin θ − cos θ|

Recall that sin θ = t, cos θ =√

1 − t². Therefore

Z 1

t −√

1 − t² dt = 1

2ln |t −√

1 − t²| − 1

2arcsin t + C

4. (8%) Evaluate the integral

Z dx

cos²x + a²sin²x and

Z dx

cos²x − a²sin²x, a > 0.

Sol:

Z dx

cos²(x) ± a²sin²(x) =

Z sec²(x)dx 1 ± a²tan²(x) Let t = tan(x) (跟一般 tan(θ

2) 代換觀念一樣 , 這邊 sin , cos 都有平方 , 是兩倍角)

=

Z dt

1 ± a²t² = 1 a

Z du

1 ± u², (u = at)

=













 1

atan⁻¹(u) + C = 1

atan⁻¹(a tan(x)) + C, for +a 1

atanh⁻¹(u) + C = 1

atanh⁻¹(a tan(x)) + C, for −a or 1

2aln   

1 + a tan(x) 1 − a tan(x)  

 for −a

5. (8%) Evaluate the improper integral Z ∞

0

xⁿe^−xdx, n is a positive integer.

Sol:

Let I_n= Z ∞

0

xⁿe^−xdx = lim

t→∞−xⁿe^−x|^t

0 + n Z ∞

0

xⁿ⁻¹e^−x

dx = nI_n−1 (since lim

t→∞xⁿe^−x   

t

0

= 0)

By Induction, I_n= nI_n−1 = n(n − 1)I_n−2= · · · = n!I₀ Consider I₀ =R∞

0 e^−xdx = lim

t→∞−e^−x  

t

0

= 1 Therefore, I_n= n!I₀ = n!

6. (8%) Evaluate the improper integral Z ∞

2

4x³+ x − 1

x²(x − 1)(x²+ 1)dx.

Sol:

f (x) ≡ 4x³+ x − 1

x²(x − 1)(x²+ 1) = 1

x² + 2

x − 1 +−2x + 1 x²+ 1 Z ∞

2

f (x)dx = Z ∞

2

1 x² +

Z ∞ 2

2 x − 1+

Z ∞ 2

−2x x²+ 1 +

Z ∞ 2

1 x²+ 1

= lim

t→∞[−1

t + 2 ln |t − 1| − ln |t²+ 1| + tan⁻¹t] − [−1

2 − ln 5 + tan⁻¹2]

= lim

t→∞[−1

x+ ln |(x − 1)²

x²+ 1 | + tan⁻¹x] +1

2+ ln 5 − tan⁻¹2

= π 2 + 1

2+ ln 5 − tan⁻¹2

7. (8%) Solve the differential equation (cos x)dy

dx + (sin x)y = sin²x · tan²x.

Sol:

dy

dx + tan x · y = sin x · tan³x · · ·N

1

I(x) = e^{Rtan xdx} = e−ln| cos x|

= sec x I(x) ·O

1

= sec x · y⁰ + sec x · tan x · y = tan⁴x d

dxsec x · y = tan⁴x ⇒ sec ·y = Z

tan⁴xdx

∵ Z

tan⁴x = Z

tan²x(sec²x − 1)dx = Z

tan²xd tan x − Z

tan²xdx

= 1

3tan³x − Z

(sec²x − 1)dx = 1

3tan³x − tan x + x + c

∴ y = 1

3tan²x sin x − sin x + cos x · x + c cos x

8. (8%) Find the orthogonal trajectories of the family of curves y = k

1 + x². (Discuss for the cases k = 0 and k 6= 0.)

Sol:

case1(k 6= 0):

[y = k

1 + x² ⇒ y + x²y = k ⇒ y⁰+ 2xy + x²y⁰ = 0 ⇒ y⁰ = −2xy 1 + x²] so the dy

dx for the orthogonal trajectory is [dy

dx = 1 + x² 2xy ⇒

Z

2ydy = Z 1

x + xdx ⇒ y² = ln |x| +x²

2 + C, y 6= 0]

case2(k = 0):

∵ k = 0 ∴ the function is y = 0(x-axis) so the orthogonal trajectory is x = c, c ∈R 9. (20%) A curve is defined by







x = t²− 2t y = −t³+ 3t²

, t ∈ R.

(a) Find the point at which the curve crosses itself.

(b) Find the points on C where the tangent is horizontal or vertical.

(c) Let R be the region enclosed by the loop of the curve. If R is rotated about the line x = −1, find the volume of the resulting solid.

(d) Find the centroid of R.

Sol:

(a) Let t1 6= t2 ∈ R , and (t12−2t1 ,−t13+3t12)=( t22−2t2 ,−t23+3t22)

⇒







t₁²−2t₁= t₂²−2t₂

−t₁³+3t₁²=−t₂³+3t₂²

⇒







(t₁− t₂)(t₁+ t₂) = 2(t₁− t₂))

3(t₁− t₂)(t₁+ t₂) = (t₁− t₂)(t²₁+ t₁t₂ + t²₂)

Since t₁ 6= t₂ ⇒







t₁+t₂= 2 t₁t₂=−2

⇒ t₁, t₂=1 ±√

3 So, we have the point (2,2) i.e. the intersection of the curve.

(b) (i) the point on C where the tangent line of horizontal Let dy

dt=0 ⇒ −3t²+6t = 0⇒ t=0 or 2 then we have the points (0,0) , (0,4).

(ii) the point on C where the tangent line of vertical (Sol) Let dy

dt=0⇒ 2t−2 = 0⇒t = 1 then we have the point (-1,2).

(c)

V = Z

2πr(dr)h

= 2π Z

(x + 1)(dx)2(y − 2)

= −8π

Z t=1+√ 3 t=1

(t − 1)⁴[(t − 1)²− 3]d(t − 1)

= −8π Z a=√

3 a=0

a⁴[a²− 3]da

= 432√ 3π 35 (d) A =

Z t=1+√ 3 t=1

−4(t²− 2t)(t − 1)[(t − 1)²− 3](t − 1)dt = 48√ 3 35 R =

Z t=1+√ 3 t=1

−4(t − 1)[(t − 1)² − 3](t − 1) d(t − 1) = 24√ 3 5 xc = A

R = 2

7, yc = 2

10. (16%) Let R be the region lying inside the curve r² = 2 cos θ and outside the curve r = 2−2 cos θ.

(a) Find the area of R.

(b) Find the area of the surface of the solid obtained by rotating R about the x-axis.

Sol:

2 cos θ = (2 − 2 cos θ)²

⇒ 4 cos²θ − 10 cos θ + 4 = 0

⇒ (2 cos θ − 1)(cos θ − 2) = 0

⇒ θ = ±π

3, which is the angle where two curve intersect.

(a)

R = 2 Z ^π₃

0

cos θ − 1

2(2 − 2 cos θ)²dθ

= Z ^π₃

0

−4 + 10 cos θ − 4 cos²θdθ

= Z ^π₃

0

−6 + 10 cos θ − 2 cos 2θdθ

= 9√ 3 2 − 2π (b)

A = Z ^π₃

0

2πyds = Z ^π₃

0

2πr sin θ√

˙r²+ r²dθ

= Z ^π₃

0

2π

√

2 cos θ sin θ

rsin²θ

2 cos θ + 2 cos θdθ + Z ^π₃

0

2π(2 − 2 cos θ) sin θp

(2 sin θ)²+ (2 − 2 cos θ)²dθ

= −2π Z ^π₃

0

√1 + 3 cos²θd cos θ + 2π Z ^π₃

0

(2 − 2 cos θ)³²d(2 − 2 cos θ)

= 2π

√3

Z tan θ=√ 3 tan θ=

√ 3 2

sec³θdθ + 4 5π

= 2π

√3(1

2sec θ tan θ + 1

2ln | sec θ + tan θ|)   

tan θ=√ 3 tan θ=

√3 2

+ 4 5π

= 14π 5 −

√7π 4 + π

√3ln 4 +√

√ 3 7 +√

3