M1120 Class 7 Dan Barbasch September 13, 2011 - - PowerPoint PPT Presentation

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.

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• x sec2 x dx.

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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.

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Further Examples

(a)

• x sin x dx.

(b)

• xe3x dx.

(c)

• x4 ln x dx.

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• x4 ln x dx

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Example (c)

Set u = ln x dv = x4 dx du = 1 x dx v = x5 5 . So

• x4 ln x dx = x5

5 ln x− 1 x x5 5 dx = x5 ln x 5 −1 5

• x4 dx = x5 ln x

5 −x5 25+C.

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Harder Examples

(d)

• eax sin (bx) dx.

(e)

• ln x dx.

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• ex sin x dx

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• 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.

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• 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 we move 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.

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Reduction formulas

The expression to be integrated depends on some integer n. We write the integral in terms of similar integrals, but for smaller n. (f)

• secn x dx.

(g)

• xnex dx.

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Example (f)

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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.

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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.

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• sin4 x · cos2 x dx.

Double Angle Formulas: sin2 θ = 1 − cos 2θ 2 cos2 θ = 1 + cos 2θ 2 sin2 x 2 cos2 x dx = 1 − cos 2x 2 2 · 1 + cos 2x 2

• dx.

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

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• 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

2 sin 2x + C, and

• cos2 2x dx =

1 + cos 4x 2

• dx = 1

2x + 1 8 sin 4x + C. Remains to compute

• cos3 2x dx.

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• cos3 2x dx.

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

• cos3 2x dx =
• cos3 u du

2 = 1 2

• cos3 u du.

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• cos3 x dx, odd powers

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• 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.

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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 will convert

• 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) reduce

the powers.

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• cos3 x dx, products of cosines

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

• cos3 x dx = 1

2

• cos x dx + 1

2

• cos x cos 2x dx =

=1 2 sin x + 1 2

• cos x · cos 2x dx.

We need to compute

• cos x · cos 2x dx.

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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)) .

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• cos x cos 2x dx

cos u cos v = 1 2 (cos (u + v) + cos (u − v)) .

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• cos x · cos 2x dx

cos u cos v = 1

2 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

2 sin x + 1 6 sin 3x + C. We used the relation cos(−x) = cos x. Plug this answer into the formula

• n a previous page.

Question: This answer does not seem to coincide with the previous

• ne.Why?!

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• cos3 x dx, recursion formula

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• 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.

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• 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.

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• sin4 x cos2 x dx

Remark: There are a lot of steps and calculations, the answer is supposed to be 1 16x − 1 64 sin 2x − 1 64 sin 4x + 1 192 sin 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 = 1

2 cos(−2x) + 1 2 cos 6x to get the answer above.

Check the arithmetic carefully.

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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 sin 2x sin 4x dx

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