m1120 class 7 - pi.math.cornell.edupi.math.cornell.edu/~web1120/slides/fall12/sep13.pdf · 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.
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∫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 .
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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)
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 .
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Harder Examples
(d)∫eax sin (bx) dx .
(e)∫
ln x dx .
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∫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 .
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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 .
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.
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∫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.
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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
2sin 2x + C ,
and ∫cos2 2x dx =
∫ (1 + cos 4x
2
)dx =
1
2x +
1
8sin 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
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 .
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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 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.
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∫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 .
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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 = 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?!
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∫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 .
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∫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
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