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M1120 Class 7 Dan Barbasch September 13, 2011 http://www.math.cornell.edu/ ˜ web1120/index.html Dan Barbasch () M1120 Class 7 September 13, 2011 1 / 26

# M1120 Class 7 - pi.math.cornell.edupi.math.cornell.edu/~web1120/slides/fall12/sep13.pdf · e2x cosx dx =e2x sinx 2 e2x cosx + 2 Z e2x cosx dx = = e2x sinx + 2e2x cosx 4 Z e2x cosx

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M1120 Class 7

Dan Barbasch

September 13, 2011

http://www.math.cornell.edu/˜web1120/index.html

Dan Barbasch () M1120 Class 7 September 13, 2011 1 / 26

Integration by Parts

Basic Formula:∫u dv = uv −

∫vdu.

d(uv) = udv + vdu ⇔udv = d(uv)− vdu ⇔∫

udv =

∫d(uv)−

∫vdu.

Dan Barbasch () M1120 Class 7 September 13, 2011 2 / 26

∫x sec2 x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 3 / 26

Example∫x sec2 x dx .

For∫x sec2 x dx set u = x and dv = sec2 x dx . Then

u = x dv = sec2 x dx

du = dx v = tan x .

The integration by parts formula gives∫x sec2 x dx = x tan x −

∫tan x dx = x tan x − ln | cos x |+ C .

Dan Barbasch () M1120 Class 7 September 13, 2011 4 / 26

Further Examples

(a)∫x sin x dx .

(b)∫xe3x dx .

(c)∫x4 ln x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 5 / 26

∫x4 ln x dx

Dan Barbasch () M1120 Class 7 September 13, 2011 6 / 26

Example (c)

Setu = ln x dv = x4 dx

du =1

xdx v =

x5

5.

So∫x4 ln x dx =

x5

5ln x−

∫1

x

x5

5dx =

x5 ln x

5−1

5

∫x4 dx =

x5 ln x

5−x5

25+C .

Dan Barbasch () M1120 Class 7 September 13, 2011 7 / 26

Harder Examples

(d)∫eax sin (bx) dx .

(e)∫

ln x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 8 / 26

∫ex sin x dx

Dan Barbasch () M1120 Class 7 September 13, 2011 9 / 26

∫e2x cos x dx

u = e2x dv = cos x dx

du = 2e2x dx v = sin x .

Then∫e2x cos x dx = e2x sin x−

∫(sin x)(2e2x dx) = e2x sin x−2

∫e2x sin x dx .

For the second integral,

u = e2x dv = sin x dx

du = 2e2x dx v = − cos x .

So ∫e2x sin x dx =− e2x cos x −

∫(− cos x)(2e2x dx) =

=− e2x cos x + 2

∫e2x cos x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 10 / 26

∫e2x cos x dx

Plugging into the first equation,∫e2x cos x dx =e2x sin x − 2

(−e2x cos x + 2

∫e2x cos x dx

)=

=(e2x sin x + 2e2x cos x

)− 4

∫e2x cos x dx .

This is an equation, the integral we want appears on both sides. So wemove it to the left and solve:

(1 + 4)

∫e2x cos x dx = e2x sin x + 2e2x cos x∫e2x cos x dx =

e2x sin x + 2e2x cos x

5+ C .

Dan Barbasch () M1120 Class 7 September 13, 2011 10 / 26

Reduction formulas

The expression to be integrated depends on some integer n. We write theintegral in terms of similar integrals, but for smaller n.

(f)∫

secn x dx .

(g)∫xnex dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 11 / 26

Example (f)

Dan Barbasch () M1120 Class 7 September 13, 2011 12 / 26

Example (f)

u = secn−2 x dv = sec2 x dx

du = (n − 2) secn−3 x sec x tan x dx v = tan x .∫secn x dx = secn−2 x tan x − (n − 2)

∫secn−2 x tan2 x dx =

= secn−2 x tan x − (n − 2)

∫secn−2 x

(−1 + sec2 x

)dx =

= secn−2 x tan x + (n − 2)

∫secn−2 x dx − (n − 2)

∫secn x dx

Now move the integral∫

secn x dx to the other side of the equation:

(1 + n − 2)

∫sec2 x dx = secn−2 x tan x + (n − 2)

∫secn−2 x dx

∫secn x dx =

secn−2 x tan x

n − 1+

n − 2

n − 1

∫secn−2 x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 13 / 26

Trigonometric integrals

∫sinm x cosn x dx ,

∫sin ax cos bx dx .

Example.∫

sin4 x cos2 x dx .

The other type will arise as we compute the example.

Dan Barbasch () M1120 Class 7 September 13, 2011 14 / 26

∫sin4 x · cos2 x dx .

Double Angle Formulas:

sin2 θ =1− cos 2θ

2cos2 θ =

1 + cos 2θ

2∫ (sin2 x

)2cos2 x dx =

∫ (1− cos 2x

2

)2

·(

1 + cos 2x

2

)dx .

The powers have decreased, but instead the angles have doubled.

Dan Barbasch () M1120 Class 7 September 13, 2011 15 / 26

∫sin4 x cos2 x dx

∫sin4 x cos2 x dx =

1

8

∫ (1− 2 cos 2x + cos2 2x

)(1 + cos 2x) dx =

=1

8

∫ (1− cos 2x − cos2 2x + cos3 2x

)dx =

=x

8− 1

8

∫cos 2x dx − 1

8

∫cos2 2x dx +

1

8

∫cos3 2x dx .

Then ∫cos 2x dx =

1

2sin 2x + C ,

and ∫cos2 2x dx =

∫ (1 + cos 4x

2

)dx =

1

2x +

1

8sin 4x + C .

Remains to compute

∫cos3 2x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 16 / 26

∫cos3 2x dx .

Change variables u = 2x , du = 2dx :∫cos3 2x dx =

∫cos3 u

du

2=

1

2

∫cos3 u du.

Dan Barbasch () M1120 Class 7 September 13, 2011 17 / 26

∫cos3 x dx , odd powers

Dan Barbasch () M1120 Class 7 September 13, 2011 18 / 26

∫cos3 x dx , odd powers

Apply the identity sin2 u + cos2 u = 1 :∫cos3 u du =

∫cos2 u · cos u du =

∫ (1− sin2 u

)· (cos u du) .

Change variables w = sin u so dw = cos u du :∫cos3 u du =

∫ (1− w2

)dw = w − w3

3+ C = sin u − sin3 u

3+ C =

= sin 2x − sin3 2x

3+ C .

Dan Barbasch () M1120 Class 7 September 13, 2011 18 / 26

General Strategy

1 When at least one of sin x or cos x appears to an odd power say sin x ,change variables u = cos θ, and use sin2 θ + cos2 θ = 1. This willconvert ∫

sinn x cosm x dx

to the integral of a polynomial, assuming n,m are positive integers.

2 For even powers, half angle formulas (or reduction formulas) reducethe powers.

Dan Barbasch () M1120 Class 7 September 13, 2011 19 / 26

∫cos3 x dx , products of cosines

cos3 x = cos x · cos2 x = cos x · 1 + cos 2x

2=

1

2cos x +

1

2cos x · cos 2x ,

So ∫cos3 x dx =

1

2

∫cos x dx +

1

2

∫cos x cos 2x dx =

=1

2sin x +

1

2

∫cos x · cos 2x dx .

We need to compute ∫cos x · cos 2x dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 20 / 26

Trigonometry Formulas

(1) cos (u + v) = cos u cos v − sin u sin v .

(2) cos (u − v) = cos u cos v + sin u sin v .

(3) sin (u + v) = sin u cos v + cos u sin v .

(4) sin (u − v) = sin u cos v − cos u sin v .

Solving,

(5) cos u cos v =1

2(cos (u − v) + cos (u + v)) .

(6) sin u sin v =1

2(cos (u − v)− cos (u + v)) .

(7) sin u cos v =1

2(sin (u + v) + sin (u − v)) .

Dan Barbasch () M1120 Class 7 September 13, 2011 21 / 26

∫cos x cos 2x dx

cos u cos v =1

2(cos (u + v) + cos (u − v)) .

Dan Barbasch () M1120 Class 7 September 13, 2011 22 / 26

∫cos x · cos 2x dx

cos u cos v = 12 cos(u + v) + 1

2 cos(u − v).∫cos x · cos 2x dx =

1

2

∫cos(x − 2x) dx +

1

2

∫cos(x + 2x) dx =

=1

2

∫cos(−x) dx +

1

2

∫cos 3x dx =

1

2sin x +

1

6sin 3x + C .

We used the relation cos(−x) = cos x . Plug this answer into the formulaon a previous page.

Question: This answer does not seem to coincide with the previousone.Why?!

Dan Barbasch () M1120 Class 7 September 13, 2011 23 / 26

∫cos3 x dx , recursion formula

Dan Barbasch () M1120 Class 7 September 13, 2011 24 / 26

∫cos3 x dx , recursion formula

Use integration by parts:

u = cos2 x dv = cos x dx

du = 2 cos x(− sin x dx) v = sin x∫cos3 x = sin x cos2 x −

∫(−2 sin x cos x) sin x dx =

= sin x cos2 x + 2

∫sin2 x cos x dx =

= sin x cos2 x + 2

∫ (1− cos2 x

)cos x dx =

= sin x cos2 x + 2

∫cos x dx − 2

∫cos3 dx =

= sin x cos2 x + 2 sin x − 2

∫cos3 dx .

Dan Barbasch () M1120 Class 7 September 13, 2011 24 / 26

∫cos3 x dx , recursion formula

This is the derivation of the recursion formula for∫

cosn x dx . Solving for∫cos3 x dx ,∫

cos3 x dx = sin x cos2 x + 2 sin x − 2

∫cos3 dx .

3

∫cos3 x dx = sin x cos2 x + 2 sin x + C .

∫cos3 x dx =

sin x cos2 x + 2 sin x

3+ C .

Dan Barbasch () M1120 Class 7 September 13, 2011 24 / 26

∫sin4 x cos2 x dx

Remark: There are a lot of steps and calculations, the answer is supposedto be

1

16x − 1

64sin 2x − 1

64sin 4x +

1

192sin 6x + C .

With this in mind, recall sin 2x = 2 sin x cos x . So we can write

sin4x cos2 x = (sin x cos x)2 · sin2 x =

(sin 2x

2

)2

·(

1− cos 2x

2

)=

=1

8

(1− cos 4x

2

)· (1− cos 2x) =

=1

16(1− cos 2x − cos 4x + cos 2x cos 4x) .

Now use cos 2x cos 4x = 12 cos(−2x) + 1

2 cos 6x to get the answer above.Check the arithmetic carefully.

Dan Barbasch () M1120 Class 7 September 13, 2011 25 / 26

Exercises for next time

(a)

∫sin5 x cos x dx (b)

∫sin4 x cos3 x dx (c)

∫sin8 x cos7 x dx

(d)

∫sin2 x dx (e)

∫sin4 x dx (f )

∫sin8 x cos2 x dx

(g)

∫sin 5x cos 2x dx (h)

∫ 2π

0sin 2x sin 4x dx

Dan Barbasch () M1120 Class 7 September 13, 2011 26 / 26