we use trigonometric substitutions to integrate algebraic...

Additional Techniques of Integration Trigonometric Substitutions

Trigonometric Substitutions

We use trigonometric substitutions to integrate algebraic functions thatcontain an expression of the form√

a2 + x2√x2 − a2√a2 − x2,

where a is a real number.

MAT 1001 Calculus I 1 / 31


Case 1

If an integrand contains a factor of the form√a2 + x2, then a

trigonometric substitutionx = a tan θ

should be considered. Thus the expressions a2 + x2 and dx becomes

a2 + x2 = a2 + a2 tan2 θ = a2(1 + tan2 θ) = a2 sec2 θ

anddx = a sec2 θ dθ.

To return to the original variable θ, we expect that substitutionx = a tan θ has an inverse.Thus by the definition of the function arctan, the inverse substitution ofthe substitution x = a tan θ is

θ = tan−1(xa

), −π

2< θ <

π

2.



Example 1

Use substitution x = 2 tan θ, −π/2 < θ < π/2, to evaluate∫1

x2√x2 + 4

dx.

Solution.

By the substitution

x = 2 tan θ, dx = 2 sec2 θ, −π/2 < θ < π/2,√x2 + 4 =

√(2 tan θ)2 + 4 =

√4 sec2 θ = 2| sec θ|.

Since the domain is −π/2 < θ < π/2, we may write | sec θ| = sec θ.



Solution (cont.)

Then the integral becomes∫1

x2√x2 + 4

dx =

∫1

(4 tan2 θ)(2 sec θ)2 sec2 θdθ

=1

4

∫sec θ

tan2 θdθ =

1

4

∫sin θ

cos3 θdθ

=∣∣ u=cos θdu=− sin θdθ

∣∣ = 1

4

∫−duu3

=1

4

(u−2

2

)+ c =

1

8cos−2 θ + c

=1

32(x2 + 4) + c

Here we used the substitution θ = arctan(x2

).



Case 2

If an integrand contains a factor of the form√x2 − a2, then a

trigonometric substitutionx = a sec θ

should be considered. Thus the expressions x2 − a2 and dx becomes

x2 − a2 = a2 sec2 θ − a2 = a2(sec2 θ − 1) = a2 tan2 θ

dx = a sec θ tan θdθ.

To return to the original variable θ, we expect that substitution has aninverse.Thus by the definition of the function arcsec, the inverse substitution ofthe substitution x = a sec θ is

θ = sec−1(xa

),

{0 ≤ θ < π

2 ,xa ≥ 1

π2 < θ ≤ π, x

a ≤ −1.



Example 2

Use substitution x = sec θ, where 0 ≤ θ < π/2 or π ≤ θ < 3π/2, to

evaluate

∫ √x2 − 1

x4dx.

Solution.

By the substitution

x = sec θ, dx = sec θ tan θ,√x2 − 1 =

√(sec θ)2 − 1 =

√tan2 θ = | tan θ|.

Since the domain is either 0 ≤ θ < π/2 or π ≤ θ < 3π/2, we may write| tan θ| = tan θ.



Solution (cont.)

Then the integral becomes∫ √x2 − 1

x4dx =

∫tan θ

sec4 θsec θ tan θdθ

=

∫tan2 θ

sec3 θdθ =

∫sin2 θ cos θdθ

=∣∣ u=sin θdu=cos θdθ

∣∣ = ∫ u2du

=

(u3

3

)+ c =

1

3sin3 θ + c

=1

3

(√x2 − 1

x

)3

+ c

Here we used the substitution θ = arcsecx.



Case 3

To evaluate the integrals that contain the expression√a2 − x2, we use the

trigonometric substitutionx = a sin θ.

Thus the expressions a2 − x2 and dx becomes

a2 − x2 = a2 − a2 sin2 θ = a2(1− sin2 θ) = a2 cos2 θ

dx = a cos θ dθ.

We have to return to the original variable x to finish off the calculation ofthe integral.Then we expect that the substitution x = a sin θ has an inverse. Thus bythe definition of the arcsin, the inverse substitution is

θ = sin−1(xa

),

−π2≤ θ ≤ π

2.



Example 3

Use substitution x = 3 sin θ, where −π/2 ≤ θ ≤ π/2, to evaluate∫ √9− x2x2

dx.

Solution.

By the substitution

x = sin θ, dx = cos θ,√9− x2 =

√9− (3 sin θ)2 =

√cos2 θ = | cos θ|.

Since the domain is −π/2 ≤ θ ≤ π/2, we may write | cos θ| = cos θ.



Solution (cont.)

Then the integral becomes∫ √9− x2x2

dx =

∫cos θ

sin2 θcos θdθ

=

∫cos2 θ

sin2 θdθ =

∫1− sin2 θ

sin2 θdθ

=

∫1

sin2 θ− 1dθ

= − cot θ − θ + c

= −√9− x2x

− arcsin(x3

)+ c

Here we used the substitution θ = arcsin(x3

).


Additional Techniques of Integration Partial Fractions

Partial Fractions

We integrate rational functions (ratios of polynomials) by expressing themas sums of simpler fractions, called partial fractions, that we alreadyknow how to integrate.



Example 4

Find

∫x3 + x

x− 1dx.

Solution.

Since the degree of the numerator is grater than the degree of thedenominator, we first perform the long division. This enables us to write∫

x3 + x

x− 1dx =

∫ (x2 + x+ 2 +

2

x− 1

)dx

=x3

3+x2

2+ 2x+ 2 ln |x− 1|+ c.



Example 5

Evaluate

∫x2 + 2x− 1

2x3 + 3x2 − 2xdx.

Solution.

Since the degree of the numerator is less than the degree of thedenominator, we don’t need to divide. We factor the denominator as

2x3 + 3x2 − 2x = x(2x2 + 3x− 2) = x(2x− 1)(x+ 2)

Since the denominator has three distinct linear factors, the partial fractiondecomposition of the integrand has the form

x2 + 2x− 1

x(2x− 1)(x+ 2)=A

x+

B

2x− 1+

C

x+ 2.



Solution (cont.)

To determine the values of A, B and C, we multiply both sides of thisequation by the product of the denominators, x(2x− 1)(x+ 2), obtaining

x2 + 2x− 1 = A(2x− 1)(x+ 2) +Bx(x+ 2) + Cx(2x− 1)

Expanding the right side of this equation and writing it in the standardform for polynomials, we get

x2 + 2x− 1 = (2A+B + 2C)x2 + (3A+ 2B − C)x− 2A

The polynomials in this equation are identical, so their coefficients mustbe equal. This gives the following system of equations for A, B, and C:

2A+B + 2C = 1

3A+ 2B − C = 2

−2A = −1



Solution (cont.)

Solving, we get A = 12 , B = 1

5 , and C = − 110 , and so∫

x2 + 2x− 1

2x3 + 3x2 − 2xdx =

∫ [1

2

1

x+

1

5

1

2x− 1− 1

10

1

x+ 2

]dx

=1

2ln |x|+ 1

10ln |2x− 1| − 1

10ln |x+ 2|+K.



Example 6

Find

∫dx

x2 − a2, where a 6= 0.

Solution.

The method of partial fractions gives

1

x2 − a2=

1

(x− a)(x+ a)=

A

x− a+

B

x+ a

and therefore A(x+ a) +B(x− a) = 1. We put x = a in this equationand get A = 1/(2a), if we put x = −a, we get B = −1/(2a). Thus∫

dx

x2 − a2=

1

2a

∫ [1

x− a− 1

x+ a

]dx

=1

2aln

∣∣∣∣x− ax+ a

∣∣∣∣+ c.



Example 7

Find

∫x4 − 2x2 + x+ 1

x3 − x2 − x+ 1dx.

Solution.

The first step is to divide. The result of long division is

x4 − 2x2 + x+ 1

x3 − x2 − x+ 1= x+ 1 +

4x

x3 − x2 − x+ 1.

The second step is to factor the denominator x3 − x2 − x+ 1. Since whenwe put x = 1 we get 0, we know that x− 1 is a factor and we obtain

x3 − x2 − x+ 1 = (x− 1)(x2 − 1) = (x− 1)2(x+ 1).



Solution (cont.)

Since the linear factor x− 1 occurs twice, the partial fractiondecomposition is

4x

(x− 1)2(x+ 1)=

A

x− 1+

B

(x− 1)2+

C

x+ 1.

Multiplying by the least common denominator, (x− 1)2(x+ 1), we get

4x = (A+ C)x2 + (B − 2C)x+ (−A+B + C).

Equating the coefficients and solving the linear system, we obtain A = 1,B = 2, and C = −1, so∫

x4 − 2x2 + x+ 1

x3 − x2 − x+ 1dx =

∫ [x+ 1 +

1

x− 1+

2

(x− 1)2− 1

x+ 1

]dx

=x2

2+ x− 2

x− 1+ ln

∣∣∣∣x− 1

x+ 1

∣∣∣∣+K.



Example 8

Find

∫4x2 − 3x+ 2

4x2 − 4x+ 3dx.

Solution.

Since the degree of the numerator is not less than the degree of thedenominator, we first divide and obtain

4x2 − 3x+ 2

4x2 − 4x+ 3= 1 +

x− 1

4x2 − 4x+ 3.

Notice that the quadratic equation 4x2 − 4x+ 3 is irreducible because itsdiscriminant is b2 − 4ac = −32 < 0. To integrate the given function wecomplete the square in the denominator:

4x2 − 4x+ 3 = (2x− 1)2 + 2.



Solution (cont.)

This suggest that we make the substitution u = 2x− 1. Then, du = 2dxand x = (u+ 1)/2, so∫

4x2 − 3x+ 2

4x2 − 4x+ 3dx =

∫ (1 +

x− 1

4x2 − 4x+ 3

)dx

= x+1

4

∫u− 1

u2 + 2du

= x+1

4

∫u

u2 + 2du− 1

4

∫1

u2 + 2

= x+1

8ln(u2 + 2)− 1

4

1√2arctan

(u√2

)+ c

= x+1

8ln(4x2 − 4x+ 3)

− 1

4√2arctan

(2x− 1√

2

)+ c.



Example 9

Write out the form of the partial fraction decomposition of the functionx3 + x2 + 1

x(x− 1)(x2 + x+ 1)(x2 + 1)3.

Solution.

x3 + x2 + 1

x(x− 1)(x2 + x+ 1)(x2 + 1)3

=A

x+

B

x− 1+

Cx+D

x2 + x+ 1+Ex+ F

x2 + 1+

Gx+H

(x2 + 1)2+

Ix+ J

(x2 + 1)3

Here, we found A = −1, B = 1/8, C = −1, D = −1, E = 15/8,F = −1/8, G = 3/4, H = 3/4, I = −1/2 and J = 1/2.



Example 10

Find

∫1− x+ 2x2 − x3

x(x2 + 1)2dx.

Solution.

The form of the partial fraction decomposition is

1− x+ 2x2 − x3

x(x2 + 1)2=A

x+Bx+ C

x2 + 1+

Dx+ E

(x2 + 1)2.

Multiplying by x(x2 + 1)2, we have

−x3 + 2x2 − x+ 1 = A(x2 + 1)2 + (Bx+ C)x(x2 + 1) + (Dx+ E)x

= (A+B)x4 + Cx3 + (2A+B +D)x2 + (C + E)x+A.



Solution (cont.)

If we equate the coefficients, we get the system

A+B = 0 C = −1 2A+B +D = 2 C + E = −1 A = 1

which has the solution A = 1, B = −1, C = −1, D = 1, and E = 0. Thus∫1− x+ 2x2 − x3

x(x2 + 1)2dx =

∫ (1

x− x+ 1

x2 + 1+

x

(x2 + 1)2

)dx

=

∫1

xdx−

∫x

x2 + 1dx−

∫1

x2 + 1dx+

∫x

(x2 + 1)2dx

= ln |x| − 1

2ln(x2 + 1)− arctanx− 1

2(x2 + 1)+ k.



Note 1

If the degree in the numerator in previous example had been the same asthat of the denominator, or higher, we would have had to take thepreliminary step of performing a long division. For instance,

2x3 − 11x2 − 2x+ 2

2x2 + x− 2= x− 6 +

5x− 4

(x+ 1)(2x− 1)

Note 2

If the denominator has more than two linear factors, we need to include aterm corresponding to each factor. For example,

x+ 6

x(x− 3)(4x+ 5)=A

x+

B

x− 3+

C

4x+ 5

where A, B and C are constants determined by solving a system of threeequations in the unknowns A, B and C.



Note 3

If a linear factor is repeated, we need to include extra terms in the partialfraction expression. Here’s an example:

x

(x+ 2)2(x− 1)=

A

x+ 2+

B

(x+ 2)2+

C

x− 1

Note 4

If we obtain an irreducible quadratic factor ax2 + bx+ c, where thediscriminant b2 − 4ac is negative. Then the corresponding partial fractionis of the form

Ax+B

a x2 + b x+ c

where A and B are constants to be determined. This term can beintegrated by completing the square and using the formula∫

dx

x2 + a2=

1

atan−1

(xa

)+ C.


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