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M1120 Class 8
Dan Barbasch
September 18, 2011
http://www.math.cornell.edu/˜web1120/index.html
Dan Barbasch () M1120 Class 8 September 18, 2011 1 / 13
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 8 September 18, 2011 2 / 13
∫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 8 September 18, 2011 3 / 13
∫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 8 September 18, 2011 4 / 13
∫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 8 September 18, 2011 5 / 13
∫cos3 x dx , odd powers
Dan Barbasch () M1120 Class 8 September 18, 2011 6 / 13
∫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 8 September 18, 2011 6 / 13
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 8 September 18, 2011 7 / 13
∫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 8 September 18, 2011 8 / 13
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 8 September 18, 2011 9 / 13
∫cos x cos 2x dx
cos u cos v =1
2(cos (u + v) + cos (u − v)) .
Dan Barbasch () M1120 Class 8 September 18, 2011 10 / 13
∫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 8 September 18, 2011 11 / 13
∫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) +
12 cos 6x to get the answer above.
Check the arithmetic carefully.
Dan Barbasch () M1120 Class 8 September 18, 2011 12 / 13
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 8 September 18, 2011 13 / 13
Trigonometric substitution
Changes integrals involving square roots, into trigonometric integrals ofthe type we studied. The formulas
1− sin2 θ = cos2 θ
1 + tan2 θ = sec2 θ
sec2 θ − 1 = tan2 θ
allow you to replace expressions with square roots by trigonometricfunctions.In one of the problems you also used the trick
1− sin x =1− 2 sinx
2cos
x
2= sin2
x
2− 2 sin
x
2cos
x
2+ cos2
x
2=
=(sin
x
2− cos
x
2
)2,√1− sin x =
∣∣∣sin x
2− cos
x
2
∣∣∣.We want methods to use systematically.
Dan Barbasch () M1120 Class 8 September 18, 2011 14 / 13
1
√a2 − x2 : x = a sin θ, dx = cos θ dθ, −π/2 ≤ θ ≤ π/2.
a2 − x2 = a2 − a2 sin2 θ = a2(1− sin2 θ
)= a2 cos2 θ.√
a2 − x2 = a| cos θ| = a cos θ.
Crucial fact: −π/2 ≤ θ ≤ π/2, so that cos θ ≥ 0.
2
√a2 + x2 : x = a tan θ, dx = sec2 θ dθ, −π/2 ≤ θ ≤ π/2.
a2 + x2 = a2 + a2 tan2 θ = a2(1 + tan2 θ
)= a2 sec2 θ.√
a2 + x2 = a| sec θ| = a sec θ.
Crucial fact: −π/2 ≤ θ ≤ π/2, so that cos θ ≥ 0.
3
√x2 − a2 : x = a sec θ, dx = sec θ tan θ dθ, 0 ≤ θ < π/2 if x ≥ a,
but π/2 < θ ≤ π if x < −a.x2 − a2 = a2 sec2 θ − a2 = a2
(sec2θ − 1
)= a2 tan2 θ.√
x2 − a2 = a| tan θ|.
Crucial fact : x ≥ a√
x2 − a2 = a tan θ, 0 ≤ θ < π/2
x ≤ −a√x2 − a2 = −a tan θ, π/2 < θ ≤ π
Dan Barbasch () M1120 Class 8 September 18, 2011 15 / 13
ExamplesCompute the area enclosed by the ellipse x2
a2+ y2
b2= 1
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Examples
Using symmetry, A = 4
∫ a
0
√b2 − b2
a2x2 dx .
Some algebra:√b2 − b2
a2x2 =
√a2b2 − b2x2
a2=
√b2
a2(a2 − x2) dx =
b
a
√a2 − x2
A = 4b
a
∫ a
0
√a2 − x2 dx
From the previous page, x = a sin θ, dx = a cos θ dθ,√a2 − x2 = a cos θ :
A =4b
a
∫ π/2
0(a cos θ)(a cos θ dθ) = 4ab
∫ π/2
0cos2 θ dθ =
=4ab
∫ π/2
0
1 + cos 2θ
2dθ = 4ab
∫ π/2
0
1
2dθ + 4ab
∫ π/2
0
cos 2θ
2dθ = πab
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Examples
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Examples
Compute
∫ −3
−6
1√x2 − 4
dx .
Observation: This integral is a POSITIVE number!From the previous page, x = 2 sec θ, dx = 2 sec θ tan θ. Because the limitsof integration are negative, π/2 < θ < π. So
√x2 − 4 = −2 tan θ.
The limits of integration are Arcsec −62 ≤ θ ≤ Arcsec −3
2 .
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Examples
∫ −3
−6
1√x2 − 4
dx =
∫ Arcsec(−3/2)
Arcsec(−3)
2 sec θ tan θ dθ
−2 tan θ=
=−∫ Arcsec(−3/2)
Arcsec(−3)sec θ dθ = − ln | sec θ + tan θ|
∣∣∣∣∣Arcsec(−3/2)
Arcsec(−3)
=
=− (ln | − 3/2− tanArcsec(−3/2)|+ ln | − 3 + tanArcsec(−3)|)
To evaluate, we use the fact that tan (Arcsec t) = −√t2 − 1 for t < 0.
This comes from reasoning on a right triangle with hypothenuse −t > 0and adjacent side 1. The opposite side is
√t2 − 1, and the tangent
function is (negative) −(√t2 − 1/1).∫ −3
−6
1√x2 − 4
dx = −(ln | − 3/2−
√5/3| − ln | − 3−
√8|)=
= ln(3 +√8)− ln(3/2 +
√5/3) (which is positive!)
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Examples
Easier Way: Change variables u = −x :∫ −3
−6
1√x2 − 4
dx =
∫ 6
3
1√u2 − 4
du.
Then 0 ≤ θ < π/2 and sec θ, tan θ and so on are all positive!
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Examples
Dan Barbasch () M1120 Class 8 September 18, 2011 16 / 13
Completing the square
∫1√
x2 + 2x + 6dx .
x2+ax+b = x2+ax+(a2
)2+
(b −
(a2
)2)=(x +
a
2
)2+c , c = b −
(a2
)2.
x2 + 2x + 6 = x2 + 2x + 1 + (6− 1) = (x + 1)2 + 5.
Change variables u = x + 1, du = dx :∫1√
x2 + 2x + 6dx =
∫1√
u2 + 5du.
Then apply the method on the previous slide.
Dan Barbasch () M1120 Class 8 September 18, 2011 17 / 13
Partial FractionsCompute
∫sec θ dθ=
∫dθ
cos θ=
cos θ dθ
cos2 θ=
∫cos θ dθ
1− sin2 θ
Change variables x = sin θ, dx = cos θ dθ :∫sec θ dθ =
∫dx
1− x2=
∫dx
(1− x)(1 + x)
Method of Partial Fractions. Write
1
(1− x)(1 + x)=
A
1− x+
B
1 + xA,B constants
and solve for A and B. Each integral separately is computable.Answer: A = 1/2, and B = 1/2.∫
dx
(1− x)(1 + x)=
1
2
∫dx
1− x+
1
2
∫dx
1 + x=
=− 1
2ln |1− x |+ 1
2ln |1 + x |+ C =
1
2ln
∣∣∣∣1 + x
1− x
∣∣∣∣+ C .
Dan Barbasch () M1120 Class 8 September 18, 2011 18 / 13
Partial Fractions
∫sec θ dθ =
1
2ln |1 + sin θ| − 1
2ln |1− sin θ|+ C =
1
2ln
∣∣∣∣1 + sin θ
1− sin θ
∣∣∣∣+ C =
= ln
√∣∣∣∣1 + sin θ
1− sin θ
∣∣∣∣+ C .
(The absolute values are not needed, the expressions are all positive!)
This does not look like ln | sec θ + tan θ|.
Dan Barbasch () M1120 Class 8 September 18, 2011 18 / 13
Partial Fractions
1 + sin u
1− sin u=(1 + sin u)(1 + sin u)
(1− sin u)(1 + sin u)=
1 + 2 sin u + sin2 u
1− sin2 u=
=(1 + sin u)2
1− sin2 u=
(1 + sin u)2
cos2 u.
√1 + sin u
1− sin u=
√(1 + sin u)2
cos2 u=
∣∣∣∣1 + sin u
cos u
∣∣∣∣ = ∣∣∣∣ 1
cos u+
sin u
cos u
∣∣∣∣ ==| sec u + tan u|.
Dan Barbasch () M1120 Class 8 September 18, 2011 18 / 13
Problems for next time:
Section 8.3 56, 58
Section 8.4 40, 60
Dan Barbasch () M1120 Class 8 September 18, 2011 19 / 13