Math 10860: Honors Calculus II, Spring 2021 Homework 7

Math 10860: Honors Calculus II, Spring 2021 Homework 7

This problem will start with a few intergrals, and then transition to questions about Taylor polynomials.

1. Some integrands appropriate for partial fractions. Do any two of these.

(a)

2x2+7x-1 x3-3x2+3x-1

dx.

(b)

3x2+3x+1 x3+2x2+2x+1

dx.

(c)

3x (x2+x+1)3

dx.

Solution:

(a) This integrand is equal to

2x2 + 7x - 1 A

B

C

=

+

+

(x - 1)3

x - 1 (x - 1)2 (x - 1)3

for some constants A, B, C. Multiplying both sides of the equation by (x - 1)3, we have 2x2 + 7x - 1 = A(x - 1)2 + B(x - 1) + C for x = 1. Letting x 1, we

have C = 8. The right side is then

A

B

8

Ax2 + (B - 2A)x + (A - B + 8)

+

+

=

,

x - 1 (x - 1)2 (x - 1)3

(x - 1)3

and equating coefficients with the numerator of the original expression, we get A = 2 and B - 2A = 7, so B = 11. Thus the integral is

2x2 + 7x - 1 dx =

x3 - 3x2 + 3x - 1

2

11

8

+

+

dx

x - 1 (x - 1)2 (x - 1)3

11

4

= 2 log |x - 1| - x - 1 - (x - 1)2 .

(b) The integrand is equal to

3x2 + 3x + 1

A

Bx + C

=

+

(x + 1)(x2 + x + 1) x + 1 x2 + x + 1

for some constants A, B, C. Multiplying both sides of the equatino by x + 1, we

have

3x2+3x+1 x2+x+1

=

A

+

(x

+

1)

Bx+C x2+x+1

for

x=

-1.

Letting

x

-1,

we

have

A=

1.

The right side is then

1

Bx + C (B + 1)x2 + (B + C + 1)x + (C + 1)

+

=

,

x + 1 x2 + x + 1

(x + 1)(x2 + x + 1)

and equation coefficients with the numerator of the original expression, we get B = 2 and C = 0. Thus the integral is

3x2 + 3x + 1 dx =

x3 + 2x2 + 2x + 1

1

2x

+

dx

x + 1 x2 + x + 1

= log |x + 1| +

2x + 1

1

-

dx

x2 + x + 1 x2 + x + 1

= log |x + 1| + log(x2 + x + 1) -

1

x

+

1 2

2

+

3 4

dx.

Letting

x

+

1 2

=

3 2

tan

,

so

dx

=

3 2

sec2

d,

the

rightmost

integral

is

1

23

x

+

1 2

2

+

3 4

dx =

3

d

23

=

3

23

2x + 1

= arctan ,

3

3

so we have

3x2 + 3x + 1 x3 + 2x2 + 2x +

1

dx

=

log |x

+

1|

+ log(x2

+x

+ 1)

-

23 3

arctan

2x + 1 3

.

(c) The integral is equal to

3x

3

2x + 1

3

1

dx =

dx -

dx

(x2 + x + 1)3

2 (x2 + x + 1)3

2 (x2 + x + 1)3

3

3

=-

-

4(x2 + x + 1)2 2

1

x

+

1 2

2

+

3 4

3 dx

3 = - 4(x2 + x + 1)2 - 96

1 (2x + 1)2 + 3 3 dx.

In the last integral, let 2x + 1 =

3 tan , so dx =

3 2

sec2

d.

The

integral

is

then

1

31

(2x + 1)2 + 3 3 dx = 54 sec4 d

3 =

cos4 d

54

= 3 (1 + cos 2)2 d

216

= 3 (1 + 2 cos 2 + cos2 2) d

216

3

1 + cos 4

=

1 + 2 cos 2 +

d

216

2

3

3

3

= + sin 2 + sin 4

144 216

1728

3

2x + 1

3

2x + 1

= arctan + sin 2 arctan

144

3

216

3

+

3 sin

4 arctan

2x+ 1

.

1728

3

Thus our integral is equal to

3

23

- 4(x2 + x + 1)2 - 3 arctan

2x+ 1 3

2 + sin

3

2 arctan

2x+ 1 3

1

2x + 1

+ sin 4 arctan

.

12

3

2. A pot-pourri with a (slightly non-obvious) trigonometric flavor. Do part (a) and one of the other two. (a) 1 - 4x - 2x2 dx. (b) cos x 9 + 25 sin2 x dx. (c) e4x 1 + e2x dx.

Solution:

(a) The integral is equal to 1 - 4x - 2x2 dx =

3 - 2(x + 1)2 dx.

Let 2(x + 1) = 3 sin , so dx = 3 cos d. The integral is then

2

3 - 2(x + 1)2 dx = 3 cos2 d 2

3 = (1 + cos 2) d

22

= 3

1 + sin 2

22

2

3

= ( + sin cos )

22

3 =

arcsin

2 (x + 1)

+

2

(x

+

1) 1

-

4x

-

2x2

.

22

3

3

(b) Let u = sin x, so du = cos x dx. The integral is then

cos x 9 + 25 sin2 x dx = 9 + 25u2 du.

Let

u

=

3 5

tan ,

so

du

=

3 5

sec2

d.

The

integral

equals

9

9 + 25u2 du =

sec3 d.

5

Integrating by parts, we have

sec3 d = sec sec2 d

= sec tan - tan sec tan d = sec tan - (sec2 - 1) sec d = sec tan + sec d - sec3 d = sec tan + log | sec + tan | - sec3 d,

so moving the adding sec3 d to both sides and dividing by 2, we have

sec3

d

=

1 (sec tan

+

log | sec

+

tan |),

2

so our integral is

cos x

9 + 25 sin2 x dx =

9 (sec tan + log | sec + tan |)

10

9 =

5 u

1 + 25 u2 + log

5 u+

1 + 25 u2

10 3

9

3

9

=

1 u9

+

25u2

+

9

log

5 u

+

1 9

+

25u2

2

10 3 3

1 = sin x

9 + 25 sin2 x +

9

log

5

1

sin x +

2

10 3

3

9 + 25 sin2 x .

(c) Let u = 1 + e2x, so du = 2e2x dx. We have e4x = e2xe2x = (u - 1)e2x, so the integral is

e4x 1

+

e2x

dx

=

1

(u

-

1) u

2e2x

dx

2

1

3

1

=

u 2 - u 2 du

2

1 25 23

=

u2 - u2

25 3

(1

+

e2x)

5 2

(1

+

e2x)

3 2

=

-

.

5

3

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