SYDE 112, Spring 2012, Assignment 4Solutions

Trigonometric substitution, rational integrals

1. Evaluate the following integrals using trigonometric substitution (re-member to include the constant!):

(a)

∫x2√

4− x2dx

Solution: We use the substitution x = 2 sin(θ) so that

θ = arcsin(x

2

)and

dx

dθ= 2 cos(θ) =⇒ dx = 2 cos(θ)dθ.

It follows that∫x2√

4− x2dx

=

∫4 sin2(θ)√

4− 4 sin2(θ)(2 cos(θ)dθ)

= 4

∫sin2(θ) cos(θ)√

cos2(θ)dθ

= 4

∫sin2(θ)dθ

= 2

∫(1− cos(2θ))dθ

= 2

[θ − sin(2θ)

2

]+ C

= 2 arcsin(x

2

)− sin

(arcsin

(x2

))cos(

arcsin(x

2

))+ C

= 2 arcsin(x

2

)− x√

4− x22

+ C.

(b)

∫ √3x2 − 1

xdx

Solution: We use the substitution x = 1√3

sec(θ) which gives

θ = arcsec(√

3x)

and

dx

dθ=

1√3

sec(θ) tan(θ) =⇒ dx =1√3

sec(θ) tan(θ)dθ.

We have∫ √3x2 − 1

xdx =

∫ √sec2(θ)− 11√3

sec(θ)

1√3

sec(θ) tan(θ)dθ

=

∫tan2(θ) dθ

= tan(θ)− θ + C

= tan(arcsec(√

3x))− arcsec(√

3x) + C

=√

3x2 − 1− arcsec(√

3x) + C.

(c)

∫x

x2 − 2x+ 2dx

Solution: We need to complete the square in the denominator.We have

x2 − 2x+ 2 = (x2 − 2x+ 1)− 1 + 2 = (x− 1)2 − 1.

It follows that we have∫x

x2 − 2x+ 2dx =

∫x

(x− 1)2 + 1dx.

We use the substitution x− 1 = tan(θ) which gives

θ = arctan(x− 1) and x = tan(θ) + 1

and

dx

dθ= tan(θ) sec(θ) =⇒ dx = tan(θ) sec(θ)dθ.

It follows that∫x

(x− 1)2 + 1dx =

∫tan(θ) + 1

sec2(θ)sec2(θ) dθ

=

∫(tan(θ) + 1) dθ

= − ln | cos(θ)|+ θ + C

= ln | cos(arctan(x− 1))|+ arctan(x− 1) + C

= ln(√

(x− 1)2 + 1) + arctan(x− 1) + C

=1

2ln((x− 1)2 + 1) + arctan(x− 1) + C.

2. Evaluate the following rational integrals using partial fraction decom-position (remember to include the constant!):

(a)

∫x

x2 − 7x+ 12dx

Solution: We have x2 − 7x + 12 = (x − 3)(x − 4). Performingthe partial fraction decomposition gives

x

x2 − 7x+ 12=

1

(x− 3)(x− 4)=

A

x− 3+

B

x− 4.

It follows thatx = A(x− 4) +B(x− 3).

Substituting x = 3 we get A = −3, and substituting x = 4 we getB = 4. It follows that we have∫

x

(x− 3)(x− 4)dx = −

∫3

x− 3dx+

∫4

x− 4dx

= 4 ln |x− 4| − 3 ln |x− 3|+ C.

(b)

∫3x2 + 3x+ 2

x3 + x2 + x+ 1dx

Solution: The term x3 + x2 + x+ 1 has x = −1 as a factor. Bysynthetic division, we have

−1 1 1 1 1−1 0 −1

1 0 1 0

It follows that x3 +x2 +x+ 1 = (x+ 1)(x2 + 1). The term x2 + 1is irreducible so that we have

3x2 + 3x+ 2

x3 + x2 + x+ 1=

3x2 + 3x+ 2

(x+ 1)(x2 + 1)=

A

x+ 1+Bx+ C

x2 + 1.

We can expand and simplify this expression to get

(3−A−B)x2 + (3−B − C)x+ (2−A− C) = 0.

This gives the system of equations

A + B = 3B + C = 3

A + C = 2

This has the solution A = C = 1 and B = 2. It follows that∫3x2 + 3x+ 2

x3 + x2 + x+ 1dx =

∫1

x+ 1dx+

∫2x+ 1

x2 + 1dx

=

∫1

x+ 1dx+

∫2x

x2 + 1dx+

∫1

x2 + 1dx

= ln |x+ 1|+ ln |x2 + 1|+ arctan(x).

3. Evaluate the definite integrals

∫ 6

4

√x2 − 4x dx

Solution: We can factor this term so that

x2 − 4x = x2 − 4x+ 4− 4 = (x− 2)2 − 4.

It follows that we have∫ 6

4

√x2 − 4x dx =

∫ 6

4

√(x− 2)2 − 4 dx.

This requires the secant substitution

x− 2 = 2 sec(x)

which gives

dx

dθ= 2 sec(θ) tan(θ) =⇒ dx = 2 sec(θ) tan(θ)dθ

and

θ = arcsec

(x− 2

2

).

The bounds also change. We have x = 4 implies θ = arcsec(1) = 0and x = 6 implies θ = arcsec(2) = π/3. It follows that∫ 6

4

√(x− 2)2 − 4 dx

=

∫ π/3

04√

sec2(θ)− 1 sec(θ) tan(θ) dθ

= 4

∫ π/3

0sec(θ) tan2(θ) dθ

= 2 [sec(θ) tan(θ)− ln | sec(θ) + tan(θ)|]π/30

= 2[2√

3− ln(2 +√

3)]

= 4√

3− 2 ln(2 +√

3).

4. Use trigonometric substitution to determine the area of the ellipsex2

a2+y2

b2= 1.

Solution: We have

x2

a2+y2

b2= 1 =⇒ y = b

√1− x2

a2.

Since the area of the portion of the ellipse in each quadrant is the samewe can consider just the area in the first quadrant and multiply by four.This gives

Area = 4b

∫ a

0

√1− x2

a2dx.

This calls for a sine substitution, so we have

x = a sin(θ) =⇒ θ = arcsin(xa

).

This givesdx

dθ= a cos(θ) =⇒ dx = a cos(θ)dθ.

We have

Area = 4b

∫ x=a

x=0

√1− sin2(θ)(a cos(θ)dθ)

= 4ab

∫ x=a

x=0cos2(θ) dθ

= 2ab

∫ x=a

x=0(1 + cos(2θ)) dθ

= 2ab

[θ +

sin(2θ)

2

]x=ax=0

= 2ab [θ + sin(θ) cos(θ)]x=ax=0

= 2ab[arcsin

(xa

)+ sin

(arcsin

(xa

))cos(

arcsin(xa

))]a0

= 2ab

[arcsin

(xa

)+(xa

) √a2 − x2a

]a0

= 2ab [arcsin(1)− arcsin(0)]

= abπ.