3.2 Trigonometric Integrals - Calculus Volume 2 | OpenStax (2024)

Table of Contents
Learning Objectives Integrating Products and Powers of sinx and cosx Example 3.8 Integrating ∫cosjxsinxdx∫cosjxsinxdx Solution Checkpoint 3.5 Example 3.9 A Preliminary Example: Integrating ∫cosjxsinkxdx∫cosjxsinkxdx Where k is Odd Solution Checkpoint 3.6 Example 3.10 Integrating an Even Power of sinxsinx Solution Checkpoint 3.7 Problem-Solving Strategy Integrating Products and Powers of sin x and cos x Example 3.11 Integrating ∫cosjxsinkxdx∫cosjxsinkxdx where k is Odd Solution Example 3.12 Integrating ∫cosjxsinkxdx∫cosjxsinkxdx where k and j are Even Solution Checkpoint 3.8 Checkpoint 3.9 Rule: Integrating Products of Sines and Cosines of Different Angles Example 3.13 Evaluating ∫sin(ax)cos(bx)dx∫sin(ax)cos(bx)dx Solution Checkpoint 3.10 Integrating Products and Powers of tanx and secx Example 3.14 Evaluating ∫secjxtanxdx∫secjxtanxdx Solution Media Checkpoint 3.11 Problem-Solving Strategy Integrating ∫tankxsecjxdx∫tankxsecjxdx Example 3.15 Integrating ∫tankxsecjxdx∫tankxsecjxdx when jj is Even Solution Example 3.16 Integrating ∫tankxsecjxdx∫tankxsecjxdx when kk is Odd Solution Example 3.17 Integrating ∫tankxdx∫tankxdx where kk is Odd and k≥3k≥3 Solution Example 3.18 Integrating ∫sec3xdx∫sec3xdx Solution Checkpoint 3.12 Reduction Formulas Rule: Reduction Formulas for ∫ sec n x d x ∫ sec n x d x and ∫ tan n x d x ∫ tan n x d x Example 3.19 Revisiting ∫sec3xdx∫sec3xdx Solution Example 3.20 Using a Reduction Formula Solution Checkpoint 3.13 Section 3.2 Exercises

Learning Objectives

  • 3.2.1Solve integration problems involving products and powers of sin x sin x and cos x . cos x .
  • 3.2.2Solve integration problems involving products and powers of tan x tan x and sec x . sec x .
  • 3.2.3Use reduction formulas to solve trigonometric integrals.

In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution, which is featured in Trigonometric Substitution. This technique allows us to convert algebraic expressions that we may not be able to integrate into expressions involving trigonometric functions, which we may be able to integrate using the techniques described in this section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let’s begin our study with products of sinxsinx and cosx.cosx.

Integrating Products and Powers of sinx and cosx

A key idea behind the strategy used to integrate combinations of products and powers of sinxsinx and cosxcosx involves rewriting these expressions as sums and differences of integrals of the form sinjxcosxdxsinjxcosxdx or cosjxsinxdx.cosjxsinxdx. After rewriting these integrals, we evaluate them using u-substitution. Before describing the general process in detail, let’s take a look at the following examples.

Example 3.8

Integrating cosjxsinxdxcosjxsinxdx

Evaluate cos3xsinxdx.cos3xsinxdx.

Solution

Use uu-substitution and let u=cosx.u=cosx. In this case, du=sinxdx.du=sinxdx. Thus,

cos 3 x sin x d x = u 3 d u = 1 4 u 4 + C = 1 4 cos 4 x + C . cos 3 x sin x d x = u 3 d u = 1 4 u 4 + C = 1 4 cos 4 x + C .

Checkpoint 3.5

Evaluate sin4xcosxdx.sin4xcosxdx.

Example 3.9

A Preliminary Example: Integrating cosjxsinkxdxcosjxsinkxdx Where k is Odd

Evaluate cos2xsin3xdx.cos2xsin3xdx.

Solution

To convert this integral to integrals of the form cosjxsinxdx,cosjxsinxdx, rewrite sin3x=sin2xsinxsin3x=sin2xsinx and make the substitution sin2x=1cos2x.sin2x=1cos2x. Thus,

cos 2 x sin 3 x d x = cos 2 x ( 1 cos 2 x ) sin x d x Let u = cos x ; then d u = sin x d x . = u 2 ( 1 u 2 ) d u = ( u 4 u 2 ) d u = 1 5 u 5 1 3 u 3 + C = 1 5 cos 5 x 1 3 cos 3 x + C . cos 2 x sin 3 x d x = cos 2 x ( 1 cos 2 x ) sin x d x Let u = cos x ; then d u = sin x d x . = u 2 ( 1 u 2 ) d u = ( u 4 u 2 ) d u = 1 5 u 5 1 3 u 3 + C = 1 5 cos 5 x 1 3 cos 3 x + C .

Checkpoint 3.6

Evaluate cos3xsin2xdx.cos3xsin2xdx.

In the next example, we see the strategy that must be applied when there are only even powers of sinxsinx and cosx.cosx. For integrals of this type, the identities

sin2x=1212cos(2x)=1cos(2x)2sin2x=1212cos(2x)=1cos(2x)2

and

cos2x=12+12cos(2x)=1+cos(2x)2cos2x=12+12cos(2x)=1+cos(2x)2

are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity cos(2x)=cos2xsin2xcos(2x)=cos2xsin2x and the Pythagorean identity cos2x+sin2x=1.cos2x+sin2x=1.

Example 3.10

Integrating an Even Power of sinxsinx

Evaluate sin2xdx.sin2xdx.

Solution

To evaluate this integral, let’s use the trigonometric identity sin2x=1212cos(2x).sin2x=1212cos(2x). Thus,

sin 2 x d x = ( 1 2 1 2 cos ( 2 x ) ) d x = 1 2 x 1 4 sin ( 2 x ) + C . sin 2 x d x = ( 1 2 1 2 cos ( 2 x ) ) d x = 1 2 x 1 4 sin ( 2 x ) + C .

Checkpoint 3.7

Evaluate cos2xdx.cos2xdx.

The general process for integrating products of powers of sinxsinx and cosxcosx is summarized in the following set of guidelines.

Problem-Solving Strategy

Integrating Products and Powers of sin x and cos x

To integrate cosjxsinkxdxcosjxsinkxdx use the following strategies:

  1. If kk is odd, rewrite sinkx=sink1xsinxsinkx=sink1xsinx and use the identity sin2x=1cos2xsin2x=1cos2x to rewrite sink1xsink1x in terms of cosx.cosx. Integrate using the substitution u=cosx.u=cosx. This substitution makes du=sinxdx.du=sinxdx.
  2. If jj is odd, rewrite cosjx=cosj1xcosxcosjx=cosj1xcosx and use the identity cos2x=1sin2xcos2x=1sin2x to rewrite cosj1xcosj1x in terms of sinx.sinx. Integrate using the substitution u=sinx.u=sinx. This substitution makes du=cosxdx.du=cosxdx. (Note: If both jj and kk are odd, either strategy 1 or strategy 2 may be used.)
  3. If both jj and kk are even, use sin2x=(1/2)(1/2)cos(2x)sin2x=(1/2)(1/2)cos(2x) and cos2x=(1/2)+(1/2)cos(2x).cos2x=(1/2)+(1/2)cos(2x). After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.

Example 3.11

Integrating cosjxsinkxdxcosjxsinkxdx where k is Odd

Evaluate cos8xsin5xdx.cos8xsin5xdx.

Solution

Since the power on sinxsinx is odd, use strategy 1. Thus,

cos 8 x sin 5 x d x = cos 8 x sin 4 x sin x d x Break off sin x . = cos 8 x ( sin 2 x ) 2 sin x d x Rewrite sin 4 x = ( sin 2 x ) 2 . = cos 8 x ( 1 cos 2 x ) 2 sin x d x Substitute sin 2 x = 1 cos 2 x . = u 8 ( 1 u 2 ) 2 ( d u ) Let u = cos x and d u = sin x d x . = ( u 8 + 2 u 10 u 12 ) d u Expand . = 1 9 u 9 + 2 11 u 11 1 13 u 13 + C Evaluate the integral . = 1 9 cos 9 x + 2 11 cos 11 x 1 13 cos 13 x + C . Substitute u = cos x . cos 8 x sin 5 x d x = cos 8 x sin 4 x sin x d x Break off sin x . = cos 8 x ( sin 2 x ) 2 sin x d x Rewrite sin 4 x = ( sin 2 x ) 2 . = cos 8 x ( 1 cos 2 x ) 2 sin x d x Substitute sin 2 x = 1 cos 2 x . = u 8 ( 1 u 2 ) 2 ( d u ) Let u = cos x and d u = sin x d x . = ( u 8 + 2 u 10 u 12 ) d u Expand . = 1 9 u 9 + 2 11 u 11 1 13 u 13 + C Evaluate the integral . = 1 9 cos 9 x + 2 11 cos 11 x 1 13 cos 13 x + C . Substitute u = cos x .

Example 3.12

Integrating cosjxsinkxdxcosjxsinkxdx where k and j are Even

Evaluate sin4xdx.sin4xdx.

Solution

Since the power on sinxsinx is even (k=4)(k=4) and the power on cosxcosx is even (j=0),(j=0), we must use strategy 3. Thus,

sin 4 x d x = ( sin 2 x ) 2 d x Rewrite sin 4 x = ( sin 2 x ) 2 . = ( 1 2 1 2 cos ( 2 x ) ) 2 d x Substitute sin 2 x = 1 2 1 2 cos ( 2 x ) . = ( 1 4 1 2 cos ( 2 x ) + 1 4 cos 2 ( 2 x ) ) d x Expand ( 1 2 1 2 cos ( 2 x ) ) 2 . = ( 1 4 1 2 cos ( 2 x ) + 1 4 ( 1 2 + 1 2 cos ( 4 x ) ) d x . sin 4 x d x = ( sin 2 x ) 2 d x Rewrite sin 4 x = ( sin 2 x ) 2 . = ( 1 2 1 2 cos ( 2 x ) ) 2 d x Substitute sin 2 x = 1 2 1 2 cos ( 2 x ) . = ( 1 4 1 2 cos ( 2 x ) + 1 4 cos 2 ( 2 x ) ) d x Expand ( 1 2 1 2 cos ( 2 x ) ) 2 . = ( 1 4 1 2 cos ( 2 x ) + 1 4 ( 1 2 + 1 2 cos ( 4 x ) ) d x .

Since cos2(2x)cos2(2x) has an even power, substitute cos2(2x)=12+12cos(4x):cos2(2x)=12+12cos(4x):

= ( 3 8 1 2 cos ( 2 x ) + 1 8 cos ( 4 x ) ) d x Simplify . = 3 8 x 1 4 sin ( 2 x ) + 1 32 sin ( 4 x ) + C Evaluate the integral . = ( 3 8 1 2 cos ( 2 x ) + 1 8 cos ( 4 x ) ) d x Simplify . = 3 8 x 1 4 sin ( 2 x ) + 1 32 sin ( 4 x ) + C Evaluate the integral .

Checkpoint 3.8

Evaluate cos3xdx.cos3xdx.

Checkpoint 3.9

Evaluate cos2(3x)dx.cos2(3x)dx.

In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include sin(ax),sin(ax), sin(bx),sin(bx), cos(ax),cos(ax), and cos(bx).cos(bx). These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.

Rule: Integrating Products of Sines and Cosines of Different Angles

To integrate products involving sin(ax),sin(ax), sin(bx),sin(bx), cos(ax),cos(ax), and cos(bx),cos(bx), use the substitutions

sin(ax)sin(bx)=12cos((ab)x)12cos((a+b)x)sin(ax)sin(bx)=12cos((ab)x)12cos((a+b)x)

(3.3)

sin(ax)cos(bx)=12sin((ab)x)+12sin((a+b)x)sin(ax)cos(bx)=12sin((ab)x)+12sin((a+b)x)

(3.4)

cos(ax)cos(bx)=12cos((ab)x)+12cos((a+b)x)cos(ax)cos(bx)=12cos((ab)x)+12cos((a+b)x)

(3.5)

These formulas may be derived from the sum-of-angle formulas for sine and cosine.

Example 3.13

Evaluating sin(ax)cos(bx)dxsin(ax)cos(bx)dx

Evaluate sin(5x)cos(3x)dx.sin(5x)cos(3x)dx.

Solution

Apply the identity sin(5x)cos(3x)=12sin(2x)+12sin(8x).sin(5x)cos(3x)=12sin(2x)+12sin(8x). Thus,

sin ( 5 x ) cos ( 3 x ) d x = 1 2 sin ( 2 x ) dx + 1 2 sin ( 8 x ) dx = 1 4 cos ( 2 x ) 1 16 cos ( 8 x ) + C . sin ( 5 x ) cos ( 3 x ) d x = 1 2 sin ( 2 x ) dx + 1 2 sin ( 8 x ) dx = 1 4 cos ( 2 x ) 1 16 cos ( 8 x ) + C .

Integrating Products and Powers of tanx and secx

Before discussing the integration of products and powers of tanxtanx and secx,secx, it is useful to recall the integrals involving tanxtanx and secxsecx we have already learned:

  1. sec2xdx=tanx+Csec2xdx=tanx+C
  2. secxtanxdx=secx+Csecxtanxdx=secx+C
  3. tanxdx=ln|secx|+Ctanxdx=ln|secx|+C
  4. secxdx=ln|secx+tanx|+C.secxdx=ln|secx+tanx|+C.

For most integrals of products and powers of tanxtanx and secx,secx, we rewrite the expression we wish to integrate as the sum or difference of integrals of the form tanjxsec2xdxtanjxsec2xdx or secjxtanxdx.secjxtanxdx. As we see in the following example, we can evaluate these new integrals by using u-substitution.

Example 3.14

Evaluating secjxtanxdxsecjxtanxdx

Evaluate sec5xtanxdx.sec5xtanxdx.

Solution

Start by rewriting sec5xtanxsec5xtanx as sec4xsecxtanx.sec4xsecxtanx.

sec 5 x tan x d x = sec 4 x sec x tan x d x Let u = sec x ; then , d u = sec x tan x d x . = u 4 d u Evaluate the integral . = 1 5 u 5 + C Substitute sec x = u . = 1 5 sec 5 x + C sec 5 x tan x d x = sec 4 x sec x tan x d x Let u = sec x ; then , d u = sec x tan x d x . = u 4 d u Evaluate the integral . = 1 5 u 5 + C Substitute sec x = u . = 1 5 sec 5 x + C

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Checkpoint 3.11

Evaluate tan5xsec2xdx.tan5xsec2xdx.

We now take a look at the various strategies for integrating products and powers of secxsecx and tanx.tanx.

Problem-Solving Strategy

Integrating tankxsecjxdxtankxsecjxdx

To integrate tankxsecjxdx,tankxsecjxdx, use the following strategies:

  1. If jj is even and j2,j2, rewrite secjx=secj2xsec2xsecjx=secj2xsec2x and use sec2x=tan2x+1sec2x=tan2x+1 to rewrite secj2xsecj2x in terms of tanx.tanx. Let u=tanxu=tanx and du=sec2xdx.du=sec2xdx.
  2. If kk is odd and j1,j1, rewrite tankxsecjx=tank1xsecj1xsecxtanxtankxsecjx=tank1xsecj1xsecxtanx and use tan2x=sec2x1tan2x=sec2x1 to rewrite tank1xtank1x in terms of secx.secx. Let u=secxu=secx and du=secxtanxdx.du=secxtanxdx. (Note: If jj is even and kk is odd, then either strategy 1 or strategy 2 may be used.)
  3. If kk is odd where k3k3 and j=0,j=0, rewrite tankx=tank2xtan2x=tank2x(sec2x1)=tank2xsec2xtank2x.tankx=tank2xtan2x=tank2x(sec2x1)=tank2xsec2xtank2x. It may be necessary to repeat this process on the tank2xtank2x term.
  4. If kk is even and jj is odd, then use tan2x=sec2x1tan2x=sec2x1 to express tankxtankx in terms of secx.secx. Use integration by parts to integrate odd powers of secx.secx.

Example 3.15

Integrating tankxsecjxdxtankxsecjxdx when jj is Even

Evaluate tan6xsec4xdx.tan6xsec4xdx.

Solution

Since the power on secxsecx is even, rewrite sec4x=sec2xsec2xsec4x=sec2xsec2x and use sec2x=tan2x+1sec2x=tan2x+1 to rewrite the first sec2xsec2x in terms of tanx.tanx. Thus,

tan 6 x sec 4 x d x = tan 6 x ( tan 2 x + 1 ) sec 2 x d x Let u = tan x and d u = sec 2 x dx . = u 6 ( u 2 + 1 ) d u Expand . = ( u 8 + u 6 ) d u Evaluate the integral . = 1 9 u 9 + 1 7 u 7 + C Substitute tan x = u . = 1 9 tan 9 x + 1 7 tan 7 x + C . tan 6 x sec 4 x d x = tan 6 x ( tan 2 x + 1 ) sec 2 x d x Let u = tan x and d u = sec 2 x dx . = u 6 ( u 2 + 1 ) d u Expand . = ( u 8 + u 6 ) d u Evaluate the integral . = 1 9 u 9 + 1 7 u 7 + C Substitute tan x = u . = 1 9 tan 9 x + 1 7 tan 7 x + C .

Example 3.16

Integrating tankxsecjxdxtankxsecjxdx when kk is Odd

Evaluate tan5xsec3xdx.tan5xsec3xdx.

Solution

Since the power on tanxtanx is odd, begin by rewriting tan5xsec3x=tan4xsec2xsecxtanx.tan5xsec3x=tan4xsec2xsecxtanx. Thus,

tan 5 x sec 3 x = tan 4 x sec 2 x sec x tan x . Write tan 4 x = ( tan 2 x ) 2 . tan 5 x sec 3 x d x = ( tan 2 x ) 2 sec 2 x sec x tan x d x Use tan 2 x = sec 2 x 1. = ( sec 2 x 1 ) 2 sec 2 x sec x tan x d x Let u = sec x and d u = sec x tan x d x . = ( u 2 1 ) 2 u 2 d u Expand . = ( u 6 2 u 4 + u 2 ) d u Integrate . = 1 7 u 7 2 5 u 5 + 1 3 u 3 + C Substitute sec x = u . = 1 7 sec 7 x 2 5 sec 5 x + 1 3 sec 3 x + C . tan 5 x sec 3 x = tan 4 x sec 2 x sec x tan x . Write tan 4 x = ( tan 2 x ) 2 . tan 5 x sec 3 x d x = ( tan 2 x ) 2 sec 2 x sec x tan x d x Use tan 2 x = sec 2 x 1. = ( sec 2 x 1 ) 2 sec 2 x sec x tan x d x Let u = sec x and d u = sec x tan x d x . = ( u 2 1 ) 2 u 2 d u Expand . = ( u 6 2 u 4 + u 2 ) d u Integrate . = 1 7 u 7 2 5 u 5 + 1 3 u 3 + C Substitute sec x = u . = 1 7 sec 7 x 2 5 sec 5 x + 1 3 sec 3 x + C .

Example 3.17

Integrating tankxdxtankxdx where kk is Odd and k3k3

Evaluate tan3xdx.tan3xdx.

Solution

Begin by rewriting tan3x=tanxtan2x=tanx(sec2x1)=tanxsec2xtanx.tan3x=tanxtan2x=tanx(sec2x1)=tanxsec2xtanx. Thus,

tan 3 x d x = ( tan x sec 2 x tan x ) d x = tan x sec 2 x d x tan x d x = 1 2 tan 2 x ln | sec x | + C . tan 3 x d x = ( tan x sec 2 x tan x ) d x = tan x sec 2 x d x tan x d x = 1 2 tan 2 x ln | sec x | + C .

For the first integral, use the substitution u=tanx.u=tanx. For the second integral, use the formula.

Example 3.18

Integrating sec3xdxsec3xdx

Integrate sec3xdx.sec3xdx.

Solution

This integral requires integration by parts. To begin, let u=secxu=secx and dv=sec2xdx.dv=sec2xdx. These choices make du=secxtanxdu=secxtanx and v=tanx.v=tanx. Thus,

sec 3 x d x = sec x tan x tan x sec x tan x d x = sec x tan x tan 2 x sec x d x Simplify . = sec x tan x ( sec 2 x 1 ) sec x d x Substitute tan 2 x = sec 2 x 1. = sec x tan x + sec x d x sec 3 x d x Rewrite . = sec x tan x + ln | sec x + tan x | sec 3 x d x . Evaluate sec x d x . sec 3 x d x = sec x tan x tan x sec x tan x d x = sec x tan x tan 2 x sec x d x Simplify . = sec x tan x ( sec 2 x 1 ) sec x d x Substitute tan 2 x = sec 2 x 1. = sec x tan x + sec x d x sec 3 x d x Rewrite . = sec x tan x + ln | sec x + tan x | sec 3 x d x . Evaluate sec x d x .

We now have

sec 3 x d x = sec x tan x + ln | sec x + tan x | sec 3 x d x . sec 3 x d x = sec x tan x + ln | sec x + tan x | sec 3 x d x .

Since the integral sec3xdxsec3xdx has reappeared on the right-hand side, we can solve for sec3xdxsec3xdx by adding it to both sides. In doing so, we obtain

2 sec 3 x d x = sec x tan x + ln | sec x + tan x | . 2 sec 3 x d x = sec x tan x + ln | sec x + tan x | .

Dividing by 2, we arrive at

sec 3 x d x = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + C . sec 3 x d x = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + C .

Checkpoint 3.12

Evaluate tan3xsec7xdx.tan3xsec7xdx.

Reduction Formulas

Evaluating secnxdxsecnxdx for values of nn where nn is odd requires integration by parts. In addition, we must also know the value of secn2xdxsecn2xdx to evaluate secnxdx.secnxdx. The evaluation of tannxdxtannxdx also requires being able to integrate tann2xdx.tann2xdx. To make the process easier, we can derive and apply the following power reduction formulas. These rules allow us to replace the integral of a power of secxsecx or tanxtanx with the integral of a lower power of secxsecx or tanx.tanx.

Rule: Reduction Formulas for sec n x d x sec n x d x and tan n x d x tan n x d x

secnxdx=1n1secn2xtanx+n2n1secn2xdxsecnxdx=1n1secn2xtanx+n2n1secn2xdx

(3.6)

tannxdx=1n1tann1xtann2xdxtannxdx=1n1tann1xtann2xdx

(3.7)

The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of tanx.tanx.

Example 3.19

Revisiting sec3xdxsec3xdx

Apply a reduction formula to evaluate sec3xdx.sec3xdx.

Solution

By applying the first reduction formula, we obtain

sec 3 x d x = 1 2 sec x tan x + 1 2 sec x d x = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + C . sec 3 x d x = 1 2 sec x tan x + 1 2 sec x d x = 1 2 sec x tan x + 1 2 ln | sec x + tan x | + C .

Example 3.20

Using a Reduction Formula

Evaluate tan4xdx.tan4xdx.

Solution

Applying the reduction formula for tan4xdxtan4xdx we have

tan 4 x d x = 1 3 tan 3 x tan 2 x d x = 1 3 tan 3 x ( tan x tan 0 x d x ) Apply the reduction formula to tan 2 x d x . = 1 3 tan 3 x tan x + 1 d x Simplify . = 1 3 tan 3 x tan x + x + C . Evaluate 1 d x . tan 4 x d x = 1 3 tan 3 x tan 2 x d x = 1 3 tan 3 x ( tan x tan 0 x d x ) Apply the reduction formula to tan 2 x d x . = 1 3 tan 3 x tan x + 1 d x Simplify . = 1 3 tan 3 x tan x + x + C . Evaluate 1 d x .

Checkpoint 3.13

Apply the reduction formula to sec5xdx.sec5xdx.

Section 3.2 Exercises

Fill in the blank to make a true statement.

69.

sin 2 x + _______ = 1 sin 2 x + _______ = 1

70.

sec 2 x 1 = _______ sec 2 x 1 = _______

Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power.

71.

sin 2 x = _______ sin 2 x = _______

72.

cos 2 x = _______ cos 2 x = _______

Evaluate each of the following integrals by u-substitution.

73.

sin 3 x cos x d x sin 3 x cos x d x

74.

cos x sin x d x cos x sin x d x

75.

tan 5 ( 2 x ) sec 2 ( 2 x ) d x tan 5 ( 2 x ) sec 2 ( 2 x ) d x

76.

sin 7 ( 2 x ) cos ( 2 x ) d x sin 7 ( 2 x ) cos ( 2 x ) d x

77.

tan ( x 2 ) sec 2 ( x 2 ) d x tan ( x 2 ) sec 2 ( x 2 ) d x

78.

tan 2 x sec 2 x d x tan 2 x sec 2 x d x

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.)

79.

sin 3 x d x sin 3 x d x

80.

cos 3 x d x cos 3 x d x

81.

sin x cos x d x sin x cos x d x

82.

cos 5 x d x cos 5 x d x

83.

sin 5 x cos 2 x d x sin 5 x cos 2 x d x

84.

sin 3 x cos 3 x d x sin 3 x cos 3 x d x

85.

sin x cos x d x sin x cos x d x

86.

sin x cos 3 x d x sin x cos 3 x d x

87.

sec x tan x d x sec x tan x d x

88.

tan ( 5 x ) d x tan ( 5 x ) d x

89.

tan 2 x sec x d x tan 2 x sec x d x

90.

tan x sec 3 x d x tan x sec 3 x d x

91.

sec 4 x d x sec 4 x d x

92.

cot x d x cot x d x

93.

csc x d x csc x d x

94.

tan 3 x sec x d x tan 3 x sec x d x

For the following exercises, find a general formula for the integrals.

95.

sin 2 a x cos a x d x sin 2 a x cos a x d x

96.

sin a x cos a x d x . sin a x cos a x d x .

Use the double-angle formulas to evaluate the following integrals.

97.

0 π sin 2 x d x 0 π sin 2 x d x

98.

0 π sin 4 x d x 0 π sin 4 x d x

99.

cos 2 3 x d x cos 2 3 x d x

100.

sin 2 x cos 2 x d x sin 2 x cos 2 x d x

101.

sin 2 x d x + cos 2 x d x sin 2 x d x + cos 2 x d x

102.

sin 2 x cos 2 ( 2 x ) d x sin 2 x cos 2 ( 2 x ) d x

For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible.

103.

0 2 π cos x sin 2 x d x 0 2 π cos x sin 2 x d x

104.

0 π sin 3 x sin 5 x d x 0 π sin 3 x sin 5 x d x

105.

0 π cos ( 99 x ) sin ( 101 x ) d x 0 π cos ( 99 x ) sin ( 101 x ) d x

106.

π π cos 2 ( 3 x ) d x π π cos 2 ( 3 x ) d x

107.

0 2 π sin x sin ( 2 x ) sin ( 3 x ) d x 0 2 π sin x sin ( 2 x ) sin ( 3 x ) d x

108.

0 4 π cos ( x / 2 ) sin ( x / 2 ) d x 0 4 π cos ( x / 2 ) sin ( x / 2 ) d x

109.

π/6π/3cos3xsinxdxπ/6π/3cos3xsinxdx (Round this answer to three decimal places.)

110.

π / 3 π / 3 sec 2 x 1 d x π / 3 π / 3 sec 2 x 1 d x

111.

0 π / 2 1 cos ( 2 x ) d x 0 π / 2 1 cos ( 2 x ) d x

112.

Find the area of the region bounded by the graphs of the equations y=sinx,y=sin3x,x=0,andx=π2.y=sinx,y=sin3x,x=0,andx=π2.

113.

Find the area of the region bounded by the graphs of the equations y=cos2x,y=sin2x,x=π4,andx=π4.y=cos2x,y=sin2x,x=π4,andx=π4.

114.

A particle moves in a straight line with the velocity function v(t)=sin(ωt)cos2(ωt).v(t)=sin(ωt)cos2(ωt). Find its position function x=f(t)x=f(t) if f(0)=0.f(0)=0.

115.

Find the average value of the function f(x)=sin2xcos3xf(x)=sin2xcos3x over the interval [π,π].[π,π].

For the following exercises, solve the differential equations.

116.

dydx=sin2x.dydx=sin2x. The curve passes through point (0,0).(0,0).

117.

d y d θ = sin 4 ( π θ ) d y d θ = sin 4 ( π θ )

118.

Find the length of the curve y=ln(cscx),π4xπ2.y=ln(cscx),π4xπ2.

119.

Find the length of the curve y=ln(sinx),π3xπ2.y=ln(sinx),π3xπ2.

120.

Find the volume generated by revolving the curve y=cos(3x)y=cos(3x) about the x-axis, 0xπ36.0xπ36.

For the following exercises, use this information: The inner product of two functions f and g over [a,b][a,b] is defined by f(x)·g(x)=f,g=abf·gdx.f(x)·g(x)=f,g=abf·gdx. Two distinct functions f and g are said to be orthogonal if f,g=0.f,g=0.

121.

Show that {sin(2x),cos(3x)}{sin(2x),cos(3x)} are orthogonal over the interval [π,π].[π,π].

122.

Evaluate ππsin(mx)cos(nx)dx.ππsin(mx)cos(nx)dx.

123.

Integrate y=tanxsec4x.y=tanxsec4x.

For each pair of integrals, determine which one is more difficult to evaluate. Explain your reasoning.

124.

sin456xcosxdxsin456xcosxdx or sin2xcos2xdxsin2xcos2xdx

125.

tan350xsec2xdxtan350xsec2xdx or tan350xsecxdxtan350xsecxdx

3.2 Trigonometric Integrals - Calculus Volume 2 | OpenStax (2024)
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