Unit 5. Integration techniques

[Pages:17]Unit 5. Integration techniques

5A. Inverse trigonometric functions; Hyperbolic functions

5A-1

a)

tan-1

3

=

b) sin-1( 3 ) =

3

23

c)tan = 5 implies sin = 5/ 26, cos = 1/ 26, cot = 1/5, csc = 26/5,

sec = 26 (from triangle)

d) sin-1 cos( ) = sin-1( 3 ) =

6

23

e) tan-1 tan( ) = 33

f)

tan-1

tan(

2 3

)

=

tan-1

tan(

- 3

)

=

- 3

g)

lim tan-1

x-

x

=

- 2

.

5A-2 a)

2 1

dx x2 +

1

=

tan-1

x

2 1

=

tan-1 2 -

4

b)

2b b

dx x2 + b2

=

2b b

d(by) (by)2 + b2

(put

x = by)

=

c)

1 -1

dx 1 - x2

=

sin-1 x

1 -1

=

2

-

- 2

=

2 1

dy b(y2 +

1)

=

1 (tan-1 b

2

-

4

)

5A-3 a) y = x - 1 , so 1 - y2 = 4x/(x + 1)2, and 1 = (x + 1) . Hence

x+1

1 - y2 2 x

dy

2

dx = (x + 1)2

d dx

sin-1

y

=

dy/dx 1 - y2

=

2 (x + 1)2

?

(x + 1) 2x

=

(x

1 + 1) x

b) sech2x = 1/ cosh2 x = 4/(ex + e-x)2

c) y = x + x2 + 1, dy/dx = 1 + x/ x2 + 1.

d ln y = dy/dx = 1 + x/ x2 + 1 = 1

dx

y

x + x2 + 1

x2 + 1

d) cos y = x = (- sin y)(dy/dx) = 1

dy dx

=

-1 sin y

=

-1 1 - x2

e) Chain rule:

d sin-1(x/a) = dx

1

?1= 1

1 - (x/a)2 a

a2 - x2

S. SOLUTIONS TO 18.01 EXERCISES

f) Chain rule:

d dx

sin-1(a/x)

=

1 1 - (a/x)2

?

-a x2

=

-a x x2 -

a2

g)

y

=

x/ 1 - x2,

dy/dx

=

(1 - x2)-3/2,

1 + y2

= 1/(1 - x2).

Thus

d dx

tan-1 y

=

dy/dx 1 + y2

=

(1 -

x2 )-3/2 (1

- x2)

=

1 1 - x2

Why is this the same as the derivative of sin-1 x?

h)

y

=

1

- x,

dy/dx

=

-1/2 1 -

x,

1-

y2

=

x.

Thus,

d dx

sin-1

y

=

dy/dx =

1 - y2 2

-1 x(1 - x)

5A-4 a) y = sinh x. A tangent line through the origin has the equation y = mx. If it meets the graph at x = a, then ma = cosh(a) and m = sinh(a). Therefore, a sinh(a) = cosh(a) .

b) Take the difference:

F (a) = a sinh(a) - cosh(a)

Newton's method for finding F (a) = 0, is the iteration

an+1 = an - F (an)/F (an) = an - tanh(an) + 1/an With a1 = 1, a2 = 1.2384, a3 = 1.2009, a4 = 1.19968. A serviceable approximation is

a 1.2

(The slope is m = sinh(a) 1.5.) The functions F and y are even. By symmetry, there is another solution -a with slope - sinh a.

5A-5 a)

y = sinh x = ex - e-x 2

y = cosh x = ex + e-x 2

y = sinh x

y is never zero, so no critical points. Inflection point x = 0; slope of y is 1 there. y is an odd function, like ex/2 for x >> 0.

b) y = sinh-1 x x = sinh y. Domain is the whole x-axis.

c) Differentiate x = sinh y implicitly with respect to x:

1

=

cosh y

?

dy dx

dy = 1 =

1

dx cosh y

sinh2 y + 1

d sinh-1 x = 1

dx

x2 + 1

y = sinh x

y = sinh 1x

5. INTEGRATION TECHNIQUES

d)

dx = x2 + a2

dx a x2 + a2/a2

d(x/a) =

(x/a)2 + 1

= sinh-1(x/a) + c

5A-6 a) 1

sin d = 2/

0

b) y = 1 - x2

=

y = -x/ 1 - x2

=

1 + (y)2 = 1/(1 - x2).

Thus

ds = w(x)dx = dx/ 1 - x2.

Therefore the average is

1 1 - x2 dx

-1

1 - x2

1 dx -1 1 - x2

1

The numerator is dx = 2. To see that these integrals are the same as the ones in part

-1

(a), take x = cos (as in polar coordinates). Then dx = - sin d and the limits of integral

are from = to = 0. Reversing the limits changes the minus back to plus:

1 1 - x2 dx

= sin d

-1

1 - x2 0

1 dx

= d =

-1 1 - x2

0

(The substitution x = sin t works similarly, but the limits of integration are -/2 and /2.) c) (x = sin t, dx = cos tdt)

11 2 -1

1

-

x2dx

=

1 2

/2

cos2 tdt =

-/2

/2 1 + cos 2t

=

dt

0

2

= /4

/2

cos2 tdt

0

5B. Integration by direct substitution

Do these by guessing and correcting the factor out front. The substitution used implicitly is given alongside the answer.

5B-1

x

x2

-

1dx

=

1 (x2

-

3

1) 2

+c

(u

=

x2

-

1,

du

=

2xdx)

3

S. SOLUTIONS TO 18.01 EXERCISES

5B-2

e8xdx = 1 e8x + c (u = 8x, du = 8dx) 8

5B-3

ln xdx x

=

1 2

(ln

x)2

+

c

(u

=

ln x,

du

=

dx/x)

5B-4

cos xdx 2 + 3 sin x

=

ln(2

+ 3 sin x) 3

+

c

(u

=

2

+

3 sin x,

du

=

3

cos xdx)

5B-5

sin2

x cos xdx

=

sin x3 3

+

c

(u

=

sin x,

du

=

cos xdx)

5B-6

sin 7xdx

=

- cos 7x 7

+

c

(u

=

7x,

du

=

7dx)

5B-7

6xdx = 6 x2 + 4 + c (u = x2 + 4, du = 2xdx) x2 + 4

5B-8 Use u = cos(4x), du = -4 sin(4x)dx,

tan 4xdx =

sin(4x)dx cos(4x)

=

-du 4u

=

-

ln u 4

+

c

=

-

ln(cos 4

4x)

+

c

5B-9

ex(1

+

ex)-1/3dx

=

3 2

(1

+

ex)2/3

+

c

(u

=

1

+

ex,

du

=

exdx)

5B-10

sec

9xdx

=

1 9

ln(sec(9x)

+

tan(9x))

+

c

(u

=

9x,

du

=

9dx)

5B-11

sec2

9xdx

=

tan 9x 9

+

c

(u

=

9x,

du

=

9dx)

5B-12

xe-x2 dx = -e-x2 + c (u = x2, du = 2xdx) 2

5B-13 u = x3, du = 3x2dx implies

x2dx 1 + x6

=

du 3(1 + u2)

=

tan-1 3

u

+

c

= tan-1(x3) + c 3

/3

sin /3

5B-14

sin3 x cos xdx =

u3du (u = sin x, du = cos xdx)

0

sin 0

3/2

3/2

=

u3du = u4/4

=

9 64

0

0

5B-15

e (ln x)3/2dx

1

= x

ln e

u3/2du (u = ln x, du = dx/x)

ln 1

5. INTEGRATION TECHNIQUES

= 1 y3/2dy = (2/5)y5/2 1 = 2

0

05

5B-16

1 -1

tan-1 xdx 1 + x2

=

tan-1 1

udu (u = tan-1 x, du = dx/(1 + x2)

tan-1 (-1)

/4

u2 /4

=

udu =

-/4

2

=0

-/4

(tan x is odd and hence tan-1 x is also odd, so the integral had better be 0)

5C. Trigonometric integrals

5C-1 sin2 xdx = 1 - cos 2x dx = x - sin 2x + c

2

24

5C-2 sin3(x/2)dx = (1 - cos2(x/2)) sin(x/2)dx = -2(1 - u2)du

(put u = cos(x/2), du = (-1/2) sin(x/2)dx)

= -2u + 2u3 + c = -2 cos(x/2) + 2cos(x/2)3 + c

3

3

5C-3 sin4 xdx = cos2(2x) dx = 4

(

1

-

cos 2

2x

)2dx

=

1 - 2 cos 2x + cos2 2x

4

dx

1

+

cos 8

4x

dx

=

x 8

+

sin 4x 32

+

c

Adding together all terms:

sin4 xdx = 3x - 1 sin(2x) + 1 sin(4x) + c

84

32

5C-4 cos3(3x)dx = (1 - sin2(3x)) cos(3x)dx = 3 cos(3x)dx)

1 - u2 3 du (u = sin(3x), du =

= u - u3 + c = sin(3x) - sin(3x)3 + c

39

3

9

5C-5 sin3 x cos2 xdx = (1 - cos2 x) cos2 x sin xdx =

du = - sin xdx)

= - u3 + u5 + c = - cos x3 + cos x5 + c

35

3

5

-(1 - u2)u2dy (u = cos x,

5C-6 sec4 xdx = (1 + tan2 x) sec2 xdx = (1 + u2)du (u = tan x, du = sec2 xdx)

= u + u3 + c = tan x + tan3 x + c

3

3

5C-7

sin2(4x) cos2(4x)dx = sin2 8xdx = 4

(1 - cos 16x)dx = 1 - sin 16x + c

8

8 128

S. SOLUTIONS TO 18.01 EXERCISES

A slower way is to use

sin2(4x) cos2(4x) =

1 - cos(8x) 2

1 + cos(8x) 2

multiply out and use a similar trick to handle cos2(8x).

5C-8

tan2(ax) cos(ax)dx = =

sin2(ax) cos(ax)

dx

1 - cos2(ax) dx

cos(ax)

5C-9 5C-10

= (sec(ax) - cos(ax))dx

=

1 a

ln(sec(ax)

+

tan(ax))

-

1 a

sin(ax)

+

c

sin3 x sec2 xdx =

1 - cos2 x cos2 x sin xdx

=

-

1

- u2 u2

du

(u = cos x, du = - sin xdx)

= u + 1 + c = cos x + sec x + c u

(tan x + cot x)2dx = tan2 x + 2 + cot2 xdx = = tan x - cot x + c

sec2 x + csc2 xdx

5C-11 sin x cos(2x)dx

= sin x(2 cos2 x - 1)dx = (1 - 2u2)du (u = cos x, du - sin xdx)

= u - 2 u3 + c = cos x - 2 cos3 x + c

3

3

5C-12

0

sin x cos(2x)dx

=

cos x

-

2 3

cos3

x

0

=

-2 3

(See 27.)

5C-13 ds = 1 + (y)2dx = 1 + cot2 xdx = csc xdx.

/2

/2

arclength =

csc xdx = - ln(csc x + cot x) = ln(1 + 2)

/4

/4

/a

/a

5C-14

sin2(ax)dx =

(1/2)(1 - cos(2ax))dx = 2/2a

0

0

5. INTEGRATION TECHNIQUES

5D. Integration by inverse substitution

5D-1 Put x = a sin , dx = a cos d:

dx

1

(a2 - x2)3/2 = a2

sec2

d

=

1 a2

tan

+

c

=

x a2 a2

-

x2

+

c

5D-2 Put x = a sin , dx = a cos d:

x3dx = a3 sin3 d = a3 (1 - cos2 ) sin d a2 - x2 = a3(- cos + (1/3) cos3 ) + c

= -a2 a2 - x2 + (a2 - x2)3/2/3 + c

5D-3 By direct substitution (u = 4 + x2),

xdx 4 + x2

=

(1/2) ln(4 + x2) + c

Put x = 2 tan , dx = 2 sec2 d,

dx 1 4 + x2 = 2 d = /2 + c

In all,

(x + 1)dx 4 + x2

=

(1/2) ln(4

+ x2)

+ (1/2) tan-1(x/2)

+c

5D-4 Put x = a sinh y, dx = a cosh ydy. Since 1 + sinh2 y = cosh2 y,

a2 + x2dx = a2 cosh2 ydy = a2 (cosh(2y) - 1)dy 2

= (a2/4) sinh(2y) - a2y/2 + c = (a2/2) sinh y cosh y - a2y/2 + c

= x a2 + x2/2 - a2 sinh-1(x/a) + c

5D-5 Put x = a sin , dx = a cos d:

a2

- x2dx x2

=

cot2 d

= (csc2 - 1)d = - ln(csc + cot ) - + c

= - ln(a/x + a2 - x2/x) - sin-1(x/a) + c 5D-6 Put x = a sinh y, dx = a cosh ydy.

x2 a2 + x2dx = a4 sinh2 y cosh2 ydy

= (a4/2) sinh2(2y)dy = a4/4 (cosh(4y) - 1)dy

= (a4/16) sinh(4y) - a4y/4 + c = (a4/8) sinh(2y) cosh(2y) - a4y/4 + c = (a4/4) sinh y cosh y(cosh2 y + sinh2 y) - a4y/4 + c = (1/4)x a2 + x2(2x2 + a2) - (a4/4) sinh-1(x/a) + c

S. SOLUTIONS TO 18.01 EXERCISES

5D-7 Put x = a sec , dx = a sec tan d:

x2

- a2dx x2

=

tan2 d sec

(sec2 - 1)d

=

sec

= (sec - cos )d

= ln(sec + tan ) - sin + c

= ln(x/a + x2 - a2/a) - x2 - a2/x + c

= ln(x + x2 - a2) - x2 - a2/x + c1 (c1 = c - ln a)

5D-8 Short way: u = x2 - 9, du = 2xdx,

x x2 - 9dx = (1/3)(x2 - 9)3/2 + c direct substitution

Long way (method of this section): Put x = 3 sec , dx = 3 sec tan d.

x x2 - 9dx = 27 sec2 tan2 d

= 27 tan2 d(tan ) = 9 tan3 + c

= (1/3)(x2 - 9)3/2 + c (tan = x2 - 9/3). The trig substitution method does not lead to a dead end, but it's not always fastest.

5D-9 y = 1/x, ds = 1 + 1/x2dx, so

b

arclength =

1 + 1/x2dx

1

Put x = tan , dx = sec2 d, x2 + 1dx = x

=

sec sec2 d tan sec (1 + tan2 ) d

tan

= (csc + sec tan )d

= - ln(csc + cot ) + sec + c = - ln( x2 + 1/x + 1/x) + x2 + 1 + c = - ln( x2 + 1 + 1) + ln x + x2 + 1 + c

arclength = - ln( b2 + 1 + 1) + ln b + b2 + 1 + ln( 2 + 1) - 2

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