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MATH 203 - Summer 2014

Instructor: Patrick Collins

Worksheet 7

#1

Evaluate the following indefinite integrals.

(a) x lnxdx(b) x3 x2 x cosxdx(c)

x2 sin

2x

dx

(d) x3 e3x dxSolution

(a) Integration by parts with u lnxand dv x dxgives du 1x

dxand v x2

2. Thus,

x lnxdx x22

lnx 12x dx

x2

2lnx x2

4 C.

(b) Integration by parts with u x3 x2 xand dv cosxdxgives du 3x2 2x 1dxand v sinx. Thus,x3 x2 x cosxdx x3 x2 x sinx 3x2 2x 1 sinxdx.

For the new integral, integration by parts with u 3x2 2x 1 and dv sinxdx gives du 6x 2dx andv cosx. Thus we obtain

x3 x2 x sinx 3x2 2x 1 cosx 6x 2 cosxdx x3 x2 x sinx 3x2 2x 1 cosx 6x 2 cosxdx.

For the new integral, integrationby parts with u 6x 2 and dv cosxdxgives du 6 dxand v sinx. Thus weobtain

x3 x2 x cosx dx x3 x2 x sinx 3x2 2x 1 cosx 6x 2 sinx 6 sinxdx x3 x2 x sinx 3x2 2x 1 cosx 6x 2 sinx 6 cosx C.

(c) Integration by parts with u x2and dv sin2xdxgives du 2x dxand v 12

cos2x. Thus,x2 sin2xdx 12 x2 cos2x x cos2xdx.

For the new integral, integration by parts with u xand dv cos2xdxgives du dxand v 12

sin2x. Thus weobtain

x2 sin2xdx 12

x2 cos2x 12

x sin2x 12sin2xdx

1

2x2 cos2x 1

2 x sin2x 1

4cos2x C.

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(d) Integration by parts with u x3and dv e3x dxgives du 3x2 dxand v 1

3e3x. Thus,

x3 e3x dx 13

x3 e3x x2 e3x dx.For the new integral, integration by parts with u x2and dv e3x dxgives du 2x dxand v

1

3e3x. Thus,

x3

e3x

dx 1

3 x3

e3x

1

3 x2

e3x

2

3x e3x

dx.For the new integral, integration by parts with u xand dv e3x dxgives du dxand v

1

3e3x. Thus,

x3 e3x dx 13

x3 e3x 1

3 x2 e3x

2

3 1

3 x e3x

1

3e3x

1

3x3 e3x

1

3 x2 e3x

2

9 x e3x

2

18e3x C.

#2

Evaluate the following definite integrals.

(a) 1

2

x ex dx

(b) 0

x2 cos2xdx

(c) 1

2

x e6x dx

Solution

(a) Integration by parts with u xand dv ex dxgives du dxand v ex. Thus,

1

2

x ex dx x ex12 1

2

ex dx

2 e2 e1 ex12 2 e2 e1 e2 e1 3 e2 2 e1.

(b) Integration by parts with u x2and dv cos2xdxgives du 2x dxand v 12

sin2x. Thus,

0

x2 cos2xdx 12

x2 sin2x0

0

x sin2xdx

1

22 sin2 1

202 sin0

0

x sin2xdx

0

x sin2xdx.

Integration by parts with u xand dv sin2xdxgives du dxand v 12 cos2x. Thus,

0

x2 cos2xdx 12

x cos2x0

1

20

cos2xdx

12

cos2 120 cos0 1

2 1

2sin2x

0

1

2

1

2 1

2sin2 1

2sin0

2.

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dy

dx

x y3

x21

1y3

dy xx21

dx

y2

2

x2 1 C

y2 2 x2 1 C

y 2 x2 1 C12

.

#4

Solve the following initial value problems.

(a)dr

d

r2

, r1 2

(b)dy

dx eyx secy 1 x2, y0 0

(c) x

dy

dx

2 y

4, y1

5.

Solution

(a) The general solution:

dr

d

r2

1r2

dr 1

d

1

r ln C.

Using the initial value condition to solve for Cgives

1

2 ln1 C C,

and so our particular solution is

1

r ln 1

2

1

r

1

2 ln

r 1

1

2ln .

(c) The general solution:

x dy

dx 2 y 4

1y4

dy 2 1x

dx

ln

y 4

2 ln

x

C.

Using the initial value condition to solve for Cgives

0 ln5 4 2 ln1 C C,and so our particular solution is

lny 4 2 lnxy 4 x2

y x2 4.

4 WS7_Solutions.nb