§8.2–Trigonometric Integrals

Mark Woodard

Furman U

Fall 2010

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 1 / 14

Outline

1 Powers of sine and cosine

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 2 / 14

Powers of sine and cosine

Essential identities

Here are the basic identities that we need:

cos2(x) + sin2(x) = 1

cos2(x) =1 + cos(2x)

2

sin2(x) =1− cos(2x)

21 + tan2(x) = sec2(x)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14

Powers of sine and cosine

Essential identities

Here are the basic identities that we need:

cos2(x) + sin2(x) = 1

cos2(x) =1 + cos(2x)

2

sin2(x) =1− cos(2x)

21 + tan2(x) = sec2(x)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14

Powers of sine and cosine

Essential identities

Here are the basic identities that we need:

cos2(x) + sin2(x) = 1

cos2(x) =1 + cos(2x)

2

sin2(x) =1− cos(2x)

21 + tan2(x) = sec2(x)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14

Powers of sine and cosine

Essential identities

Here are the basic identities that we need:

cos2(x) + sin2(x) = 1

cos2(x) =1 + cos(2x)

2

sin2(x) =1− cos(2x)

2

1 + tan2(x) = sec2(x)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14

Powers of sine and cosine

Essential identities

Here are the basic identities that we need:

cos2(x) + sin2(x) = 1

cos2(x) =1 + cos(2x)

2

sin2(x) =1− cos(2x)

21 + tan2(x) = sec2(x)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 3 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 1

Consider an integral of the form

∫cosm(x) sinn(x)dx : m or n is odd.

We use the identity sin2(x) + cos2(x) = 1 and u-substitution.

Problem

Evaluate I =

∫sin4(x) cos7(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 1

Consider an integral of the form

∫cosm(x) sinn(x)dx : m or n is odd.

We use the identity sin2(x) + cos2(x) = 1 and u-substitution.

Problem

Evaluate I =

∫sin4(x) cos7(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 1

Consider an integral of the form

∫cosm(x) sinn(x)dx : m or n is odd.

We use the identity sin2(x) + cos2(x) = 1 and u-substitution.

Problem

Evaluate I =

∫sin4(x) cos7(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 1

Consider an integral of the form

∫cosm(x) sinn(x)dx : m or n is odd.

We use the identity sin2(x) + cos2(x) = 1 and u-substitution.

Problem

Evaluate I =

∫sin4(x) cos7(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 4 / 14

Powers of sine and cosine

Solution

Write

sin4(x) cos7(x) = sin4(x)(cos2(x)

)3cos(x)

= sin4(x)(1− sin2(x)

)3cos(x).

Let u = sin(x). Then du = cos(x)dx and I =

∫u4(1− u2)3du.

Then

I =

∫ (u4 − 3u6 + 3u8 − u10

)du

=u5

5− 3u7

7+

3u9

9− u11

11+ C

=sin5(x)

5− 3 sin7(x)

7+

3 sin9(x)

9− sin11(x)

11+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14

Powers of sine and cosine

Solution

Write

sin4(x) cos7(x) = sin4(x)(cos2(x)

)3cos(x)

= sin4(x)(1− sin2(x)

)3cos(x).

Let u = sin(x). Then du = cos(x)dx and I =

∫u4(1− u2)3du.

Then

I =

∫ (u4 − 3u6 + 3u8 − u10

)du

=u5

5− 3u7

7+

3u9

9− u11

11+ C

=sin5(x)

5− 3 sin7(x)

7+

3 sin9(x)

9− sin11(x)

11+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14

Powers of sine and cosine

Solution

Write

sin4(x) cos7(x) = sin4(x)(cos2(x)

)3cos(x)

= sin4(x)(1− sin2(x)

)3cos(x).

Let u = sin(x). Then du = cos(x)dx and I =

∫u4(1− u2)3du.

Then

I =

∫ (u4 − 3u6 + 3u8 − u10

)du

=u5

5− 3u7

7+

3u9

9− u11

11+ C

=sin5(x)

5− 3 sin7(x)

7+

3 sin9(x)

9− sin11(x)

11+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14

Powers of sine and cosine

Solution

Write

sin4(x) cos7(x) = sin4(x)(cos2(x)

)3cos(x)

= sin4(x)(1− sin2(x)

)3cos(x).

Let u = sin(x). Then du = cos(x)dx and I =

∫u4(1− u2)3du.

Then

I =

∫ (u4 − 3u6 + 3u8 − u10

)du

=u5

5− 3u7

7+

3u9

9− u11

11+ C

=sin5(x)

5− 3 sin7(x)

7+

3 sin9(x)

9− sin11(x)

11+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 5 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 2

Consider an integral of the form

∫cosm(x) sinn(x)dx : m and n even.

Use the the half-angle identities, repeatedly if necessary.

Problem

Solve the integral

∫ π/2

0sin2(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 2

Consider an integral of the form

∫cosm(x) sinn(x)dx : m and n even.

Use the the half-angle identities, repeatedly if necessary.

Problem

Solve the integral

∫ π/2

0sin2(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 2

Consider an integral of the form

∫cosm(x) sinn(x)dx : m and n even.

Use the the half-angle identities, repeatedly if necessary.

Problem

Solve the integral

∫ π/2

0sin2(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14

Powers of sine and cosine

Powers of sine and cosine, Case 2

Consider an integral of the form

∫cosm(x) sinn(x)dx : m and n even.

Use the the half-angle identities, repeatedly if necessary.

Problem

Solve the integral

∫ π/2

0sin2(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 6 / 14

Powers of sine and cosine

Solution

We will solve the anti-derivative problem I =

∫sin2(x)dx first and

evaluate at the end.

Since sin2(x) = (1− cos(2x))/2, we have

I =

∫1− cos(2x)

2dx =

x

2− sin(2x)

4+ C .

We can drop the “+C” and evaluate the resulting anti-derivativebetween the limits, obtaining∫ π/2

0sin2(x)dx =

π

4.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14

Powers of sine and cosine

Solution

We will solve the anti-derivative problem I =

∫sin2(x)dx first and

evaluate at the end.

Since sin2(x) = (1− cos(2x))/2, we have

I =

∫1− cos(2x)

2dx =

x

2− sin(2x)

4+ C .

We can drop the “+C” and evaluate the resulting anti-derivativebetween the limits, obtaining∫ π/2

0sin2(x)dx =

π

4.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14

Powers of sine and cosine

Solution

We will solve the anti-derivative problem I =

∫sin2(x)dx first and

evaluate at the end.

Since sin2(x) = (1− cos(2x))/2, we have

I =

∫1− cos(2x)

2dx =

x

2− sin(2x)

4+ C .

We can drop the “+C” and evaluate the resulting anti-derivativebetween the limits, obtaining∫ π/2

0sin2(x)dx =

π

4.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14

Powers of sine and cosine

Solution

We will solve the anti-derivative problem I =

∫sin2(x)dx first and

evaluate at the end.

Since sin2(x) = (1− cos(2x))/2, we have

I =

∫1− cos(2x)

2dx =

x

2− sin(2x)

4+ C .

We can drop the “+C” and evaluate the resulting anti-derivativebetween the limits, obtaining∫ π/2

0sin2(x)dx =

π

4.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 7 / 14

Powers of sine and cosine

Problem

Evaluate I =

∫sin2(x) cos2(x)dx.

Solution

By two applications of the double-angle formulas we find

sin2(x) cos2(x) =1− cos2(2x)

4

=1

8− cos(4x)

8.

Finally

I =

∫ (1

8− cos(4x)

8

)dx =

x

8− sin(4x)

32+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14

Powers of sine and cosine

Problem

Evaluate I =

∫sin2(x) cos2(x)dx.

Solution

By two applications of the double-angle formulas we find

sin2(x) cos2(x) =1− cos2(2x)

4

=1

8− cos(4x)

8.

Finally

I =

∫ (1

8− cos(4x)

8

)dx =

x

8− sin(4x)

32+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14

Powers of sine and cosine

Problem

Evaluate I =

∫sin2(x) cos2(x)dx.

Solution

By two applications of the double-angle formulas we find

sin2(x) cos2(x) =1− cos2(2x)

4

=1

8− cos(4x)

8.

Finally

I =

∫ (1

8− cos(4x)

8

)dx =

x

8− sin(4x)

32+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14

Powers of sine and cosine

Problem

Evaluate I =

∫sin2(x) cos2(x)dx.

Solution

By two applications of the double-angle formulas we find

sin2(x) cos2(x) =1− cos2(2x)

4

=1

8− cos(4x)

8.

Finally

I =

∫ (1

8− cos(4x)

8

)dx =

x

8− sin(4x)

32+ C

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 8 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 1

Consider an integral of the form

∫tanm(x) secn(x)dx where n, the

power of the secant, is even.

In this case, keep a sec2(x) and convert the remaining secants totangents through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = tan(x), du = sec2(x)dx .

Problem

Solve I =

∫ π/4

0tan4(x) sec6(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 1

Consider an integral of the form

∫tanm(x) secn(x)dx where n, the

power of the secant, is even.

In this case, keep a sec2(x) and convert the remaining secants totangents through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = tan(x), du = sec2(x)dx .

Problem

Solve I =

∫ π/4

0tan4(x) sec6(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 1

Consider an integral of the form

∫tanm(x) secn(x)dx where n, the

power of the secant, is even.

In this case, keep a sec2(x) and convert the remaining secants totangents through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = tan(x), du = sec2(x)dx .

Problem

Solve I =

∫ π/4

0tan4(x) sec6(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 1

Consider an integral of the form

∫tanm(x) secn(x)dx where n, the

power of the secant, is even.

In this case, keep a sec2(x) and convert the remaining secants totangents through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = tan(x), du = sec2(x)dx .

Problem

Solve I =

∫ π/4

0tan4(x) sec6(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 1

Consider an integral of the form

∫tanm(x) secn(x)dx where n, the

power of the secant, is even.

In this case, keep a sec2(x) and convert the remaining secants totangents through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = tan(x), du = sec2(x)dx .

Problem

Solve I =

∫ π/4

0tan4(x) sec6(x)dx.

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 9 / 14

Powers of sine and cosine

Solution

Write

tan4(x) sec6(x) = tan4(x)(sec2(x)

)2sec2(x)

= tan4(x)(1 + tan2(x)

)2sec2(x).

We make the substitution u = tan(x) and du = sec2(x)dx; thus,

I =

∫ 1

0u4(1 + u2)2du =

∫ 1

0(u4 + 2u6 + u8)du =

143

315

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 10 / 14

Powers of sine and cosine

Solution

Write

tan4(x) sec6(x) = tan4(x)(sec2(x)

)2sec2(x)

= tan4(x)(1 + tan2(x)

)2sec2(x).

We make the substitution u = tan(x) and du = sec2(x)dx; thus,

I =

∫ 1

0u4(1 + u2)2du =

∫ 1

0(u4 + 2u6 + u8)du =

143

315

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 10 / 14

Powers of sine and cosine

Solution

Write

tan4(x) sec6(x) = tan4(x)(sec2(x)

)2sec2(x)

= tan4(x)(1 + tan2(x)

)2sec2(x).

We make the substitution u = tan(x) and du = sec2(x)dx; thus,

I =

∫ 1

0u4(1 + u2)2du =

∫ 1

0(u4 + 2u6 + u8)du =

143

315

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 10 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 2

Consider an integral of the form

∫tanm(x) secn(x)dx where m, the

power of the tangent, is odd.

In this case, keep one tangent and convert the remaining tangents tosecants through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx .

Problem

Solve

∫ π/3

0tan7(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 2

Consider an integral of the form

∫tanm(x) secn(x)dx where m, the

power of the tangent, is odd.

In this case, keep one tangent and convert the remaining tangents tosecants through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx .

Problem

Solve

∫ π/3

0tan7(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 2

Consider an integral of the form

∫tanm(x) secn(x)dx where m, the

power of the tangent, is odd.

In this case, keep one tangent and convert the remaining tangents tosecants through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx .

Problem

Solve

∫ π/3

0tan7(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 2

Consider an integral of the form

∫tanm(x) secn(x)dx where m, the

power of the tangent, is odd.

In this case, keep one tangent and convert the remaining tangents tosecants through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx .

Problem

Solve

∫ π/3

0tan7(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14

Powers of sine and cosine

Powers of tangent and secant, Case 2

Consider an integral of the form

∫tanm(x) secn(x)dx where m, the

power of the tangent, is odd.

In this case, keep one tangent and convert the remaining tangents tosecants through 1 + tan2(x) = sec2(x).

Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx .

Problem

Solve

∫ π/3

0tan7(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 11 / 14

Powers of sine and cosine

Solution

Write

tan7(x) sec5(x) = tan(x)(tan2(x)

)3sec5(x)

= tan(x)(sec2(x)− 1)3 sec5(x)

If u = sec(x), then du = tan(x) sec(x)dx and

I =

∫ 2

1(u2 − 1)3u4du

=

∫ 2

1(u10 − 3u8 + 3u6 − u4)du

=4779

40

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 12 / 14

Powers of sine and cosine

Solution

Write

tan7(x) sec5(x) = tan(x)(tan2(x)

)3sec5(x)

= tan(x)(sec2(x)− 1)3 sec5(x)

If u = sec(x), then du = tan(x) sec(x)dx and

I =

∫ 2

1(u2 − 1)3u4du

=

∫ 2

1(u10 − 3u8 + 3u6 − u4)du

=4779

40

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 12 / 14

Powers of sine and cosine

Solution

Write

tan7(x) sec5(x) = tan(x)(tan2(x)

)3sec5(x)

= tan(x)(sec2(x)− 1)3 sec5(x)

If u = sec(x), then du = tan(x) sec(x)dx and

I =

∫ 2

1(u2 − 1)3u4du

=

∫ 2

1(u10 − 3u8 + 3u6 − u4)du

=4779

40

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 12 / 14

Powers of sine and cosine

A note on powers of tangents and secants

Our analysis is not exhaustive:

We do not have any direct methods for integrating powers of tangentalone and powers of secant alone, for example,∫

tan8(x)dx and

∫sec4(x)dx .

These integrals can be solved, but we will not address these cases bydirect methods.

We do not have any methods for integrating an even power oftangent times an odd power of secant, for example,∫

tan4(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 13 / 14

Powers of sine and cosine

A note on powers of tangents and secants

Our analysis is not exhaustive:

We do not have any direct methods for integrating powers of tangentalone and powers of secant alone, for example,∫

tan8(x)dx and

∫sec4(x)dx .

These integrals can be solved, but we will not address these cases bydirect methods.

We do not have any methods for integrating an even power oftangent times an odd power of secant, for example,∫

tan4(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 13 / 14

Powers of sine and cosine

A note on powers of tangents and secants

Our analysis is not exhaustive:

We do not have any direct methods for integrating powers of tangentalone and powers of secant alone, for example,∫

tan8(x)dx and

∫sec4(x)dx .

These integrals can be solved, but we will not address these cases bydirect methods.

We do not have any methods for integrating an even power oftangent times an odd power of secant, for example,∫

tan4(x) sec5(x)dx .

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 13 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint:

this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint:

this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint:

this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint:

this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint:

this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint:

use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint:

use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint:

save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint:

save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint:

try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint:

try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14

Powers of sine and cosine

Problem (Some ad hoc problems)

Show that∫

tan(x)dx = ln | sec(x)|+ C

Show that∫

sec(x)dx = ln | sec(x) + tan(x)|+ C

Evaluate∫

sec2(x)dx Hint: this is easy!

Evaluate∫

tan2(x)dx Hint: use a nice trig identity.

Evaluate∫

tan3(x)dx Hint: save out a tangent and then use a trigidentity.

Evaluate∫

sec3(x)dx Hint: try parts with u = sec x. (see page 500.)

Mark Woodard (Furman U) §8.2–Trigonometric Integrals Fall 2010 14 / 14