M1120 Class 8 - Cornell Universitypi.math.cornell.edu/~web1120/slides/fall12/sep18.pdfGeneral Strategy 1 When at least one of sinx or cosx appears to an odd powersay sinx, change variables

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.


∫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.


∫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 .


∫cos3 2x dx .

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

∫cos3 u

du

2=

1

2

∫cos3 u du.


∫cos3 x dx , odd powers


∫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 .


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.


∫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 .


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)) .


∫cos x cos 2x dx

cos u cos v =1

2(cos (u + v) + cos (u − v)) .


∫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?!


∫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.


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


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.


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


ExamplesCompute the area enclosed by the ellipse x2

a2+ y2

b2= 1


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


Examples


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 .


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!)


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!


Examples


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.


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 .


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|.


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|.


Problems for next time:

Section 8.3 56, 58

Section 8.4 40, 60


Documents

M1120 Class 8 - Cornell Universitypi.math.cornell.edu/~web1120/slides/fall12/sep18.pdfGeneral Strategy 1 When at least one of sinx or cosx appears to an odd powersay sinx, change variables