Arkansas Tech UniversityMATH 2924: Calculus II

Dr. Marcel B. Finan

Solutions to Assignment 6.2

Exercise 3Let u = cosx then du = − sinxdx. Thus,∫ π/2

0sin7 x cos5 xdx =

∫ π/2

0(1− cos2 x)3 sinx cos5 xdx

=

∫ 1

0(1− u2)3u5du =

∫ 1

0(1− 3u2 + 3u4 − u6)u5du

=

[u6

6− 3

8u8 +

3

10u10 − 1

12u12]10

=1

120

Exercise 5Using the trigonometric identity

cos2 x =1 + cos 2x

2

we find ∫ π/2

0cos2 xdx =

∫ π/2

0

1 + cos 2x

2dx

=1

2

∫ π/2

0(1 + cos 2x)dx

=1

2

[x+

sin 2x

2

]π/20

=π

4

Exercise 7Letting u = 2x and using the trigonometric identity

cos2 u =1 + cos 2u

2

1

we find∫ π

0cos4 2xdx =

1

2

∫ 2π

0[cos2 u]2du =

1

2

∫ 2π

0

(1 + cos 2u

2

)2

du

=1

8

∫ 2π

0(1 + 2 cos 2u+ cos2 2u)du

=1

8

[u+ sin 2u+

u

2+

sin 4u

8

]2π0

=3π

8

Exercise 11Using integration by parts with u = t and v′ = sin2 t, we have∫

t sin2 tdt =t

2

(t− sin 2t

2

)−∫

1

2

(t− sin 2t

2

)dt

=t

2

(t− sin 2t

2

)− t2

4− cos 2t

8+ C

=t2

4− t

4sin 2t− cos 2t

8+ C

Exercise 17Letting u = secx so that du = tanx secxdx, we have∫

tanx sec3 xdx =

∫u2du =

u3

3+ C

=sec3 x

3+ C

Exercise 21Letting u = tanx so that du = secx2dx, we have∫

tan4 x sec6 xdx =

∫u4(u2 + 1)2du

=

∫(u8 + 2u6 + u4)du =

u9

9+

2

7u7 +

u5

5+ C

=tan9 x

9+

2

7tan7 x+

tan5 x

5+ C

2

Exercise 27Letting u = tanx so that du = sec2 xdx and using Example 7 of the book,we have ∫

tan5 xdx =

∫tan3 x tan2 xdx =

∫tan3 x(sec2 x− 1)dx

=

∫tan3 x sec2 xdx−

∫tan3 xdx

=tan4 x

4− tan2 x

2+ ln | secx|+ C

Exercise 33Using the fact that

(cscx− cotx)′ = − cscx cotx+ csc2 x = cscx(cscx− cotx)

we have ∫cscxdx =

∫cscx(cscx− cotx)

cscx− cotxdx

=

∫(cscx− cotx)′

cscx− cotxdx

= ln | cscx− cotx|+ C

Exercise 39Let x = 2 sin θ with −π

2 < θ < π2 . Then

√4− x2 = 2 cos θ. We have∫

dx

x2√

4− x2=

∫2 cos θ

4 sin2 θ(2 cos θ)dθ

=1

4

∫csc2 θdθ = −1

4cot θ

=−√

4− x24x

+ C

3

Exercise 40Let x = 2 tan θ with −π

2 < θ < π2 . Then

√x2 + 4 = 2 sec θ. We have∫

x3√x2 + 4

dx =

∫8 tan3 θ(2 sec2 θ)

2 sec θdθ

=8

∫tan3 θ sec θdθ = 8

∫(sec2 θ − 1) tan θ sec θdθ

=8

∫(u2 − 1)du = 8

[sec3 θ

3− sec θ

]+ C

=1

3(x2 + 4)

√x2 + 4− 4

√x2 + 4 + C

Exercise 41Let x = 2 sec θ with 0 < θ < π

2 or π < θ < 3π2 . Then

√x2 − 4 = 2 tan θ. We

have ∫ √x2 − 4

xdx =

∫2 tan θ(2 sec θ tan θ)

2 sec θdθ

=2

∫tan2 θdθ = 2

∫(sec2 θ − 1)dθ

=2 tan θ − 2θ + C =√x2 − 4− 2 sec−1

(x2

)+ C

4

Exercise 45Let x = a tan θ with −π

2 < θ < π2 . Then

√a2 + x2 = a sec θ and dx =

a sec2 θdθ. Hence, ∫ a

0

dx

(a2 + x2)32

=

∫ π/4

0

a sec2 θ

a3 sec3 θdθ

=1

a2sin θ]

π/40 =

√2

2a2

Exercise 49Let 2x = sin θ with −π

2 < θ < π2 . Then

√1− 4x2 = cos θ and 2dx = cos θdθ.

Thus, ∫ √1− 4x2dx =

1

2

∫cos2 θdθ

=1

4

∫(1 + cos 2θ)dθ =

θ

4+

sin 2θ

8+ C

=θ

4+

2 sin θ cos θ

8+ C

=1

4sin−1 (2x) +

x√

1− 4x2

2+ C

Exercise 51Let x = 3 sec θ so that

√x2 − 9 = 3 tan θ and dx = 3 sec θ tan θdθ. Thus,∫ √

x2 − 9

x3dx =

1

3

∫tan2 θ

sec2 θdθ

=1

3

∫sin2 θdθ =

θ

6− 1

12sin 2θ + C

=1

6sec−1

(x3

)−√x2 − 9

2x+ C

5

Exercise 53Let 5x = 3 sin θ with −π

2 < θ < π2 . Then

√9− 25x2 = 3 cos θ and 5dx =

3 cos θdθ. Thus, ∫ 0.6

0

x2√9− 25x2

dx =9

125

∫ π/2

0sin2 θdθ

=9

125

[θ

2− sin 2θ

4

]π/20

=9

125[π

4] =

9π

500

Exercise 57Let x = tan θ so that

√x2 + 1 = sec θ and dx = sec2 θdθ. Thus,∫ √

x2 + 1

xdx =

∫sec3 θ

tan θdθ

=

∫ (tan2 θ + 1

tan θ

)sec θdθ

=

∫ (tan θ +

1

tan θ

)sec θdθ

=

∫sin θ

cos2 θdθ +

∫csc θdθ

= sec θ + ln | csc θ − cot θ|+ C

=√x2 + 1 + ln

∣∣∣∣∣√x2 + 1

x− 1

x

∣∣∣∣∣+ C

Exercise 65We have

f(t) =

∫ t

0sinωs cos2 ωsds = − 1

3ωcos3 ωs

∣∣∣∣t0

=1

3ω

(1− cos3 ωt

)Exercise 66(a) Recall that the average of f(x) in [a, b] is given by

1

b− a

∫ b

af(x)dx.

6

Now, one cycle takes 160 seconds. We are asked to find[

1160 − 0

∫ 1/60

0E(t)2dt

] 12

.

We have∫ 1/60

0E(t)2dt =24, 025

∫ 1/60

0sin2 (120πt)dt =

24, 025

120π

[60πt− 1

4sin (240πt)

]1/600

=24, 025

120π(π) =

4805

24.

Thus,

RMS =

[1

160 − 0

4805

24

] 12

≈ 109.60.

(b) We must solve the equation

2202 =1

160 − 0

∫ 1/60

0A2 sin2 (120πt)dt.

That is,

2202 =60A2

120

which yields A ≈ 311.13

Exercise 67Using the substitution x = sec θ, we have

1

7− 1

∫ 7

1

√x2 − 1

xdx =

1

6

∫ sec−1 7

sec−1 1tan2 θdθ

=1

6

∫ sec−1 7

sec−1 1(sec2 θ − 1)dθ

=1

6[tan θ − θ]sec

−1 7sec−1 1

=1

6

[√x2 − 1− sec−1 x

]71

=1

6(√

48− sec−1 7)

7

Exercise 70From Exercise 45, we find∫

dx

(x2 + b2)32

=sin θ

b2+ C =

x

b2√x2 + b2

+ C.

Thus,

E(P ) =λb

4πε0

[x

b2√x2 + b2

]L−a−a

=λ

4πε0b

[L− a√

(L− a)2 + b2+

a√a2 + b2

]

8