Learning Objectives
- 3.2.1 Solve integration problems involving products and powers of sinx and cosx.
- 3.2.2 Solve integration problems involving products and powers of tanx and secx.
- 3.2.3 Use 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 sinx and cosx.
Integrating Products and Powers of sinx and cosx
A key idea behind the strategy used to integrate combinations of products and powers of sinx and cosx involves rewriting these expressions as sums and differences of integrals of the form ∫sinjxcosxdx or ∫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 ∫cosjxsinxdx
Evaluate ∫cos3xsinxdx.
Solution
Use u-substitution and let u=cosx. In this case, du=−sinxdx. Thus,
Checkpoint 3.5
Evaluate ∫sin4xcosxdx.
Example 3.9
A Preliminary Example: Integrating ∫cosjxsinkxdx Where k is Odd
Evaluate ∫cos2xsin3xdx.
Solution
To convert this integral to integrals of the form ∫cosjxsinxdx, rewrite sin3x=sin2xsinx and make the substitution sin2x=1−cos2x. Thus,
Checkpoint 3.6
Evaluate ∫cos3xsin2xdx.
In the next example, we see the strategy that must be applied when there are only even powers of sinx and cosx. For integrals of this type, the identities
and
are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity cos(2x)=cos2x−sin2x and the Pythagorean identity cos2x+sin2x=1.
Example 3.10
Integrating an Even Power of sinx
Evaluate ∫sin2xdx.
Solution
To evaluate this integral, let’s use the trigonometric identity sin2x=12−12cos(2x). Thus,
Checkpoint 3.7
Evaluate ∫cos2xdx.
The general process for integrating products of powers of sinx and cosx is summarized in the following set of guidelines.
Problem-Solving Strategy
Integrating Products and Powers of sin x and cos x
To integrate ∫cosjxsinkxdx use the following strategies:
- If k is odd, rewrite sinkx=sink−1xsinx and use the identity sin2x=1−cos2x to rewrite sink−1x in terms of cosx. Integrate using the substitution u=cosx. This substitution makes du=−sinxdx.
- If j is odd, rewrite cosjx=cosj−1xcosx and use the identity cos2x=1−sin2x to rewrite cosj−1x in terms of sinx. Integrate using the substitution u=sinx. This substitution makes du=cosxdx. (Note: If both j and k are odd, either strategy 1 or strategy 2 may be used.)
- If both j and k are even, use sin2x=(1/2)−(1/2)cos(2x) and cos2x=(1/2)+(1/2)cos(2x). After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.
Example 3.11
Integrating ∫cosjxsinkxdx where k is Odd
Evaluate ∫cos8xsin5xdx.
Solution
Since the power on sinx is odd, use strategy 1. Thus,
Example 3.12
Integrating ∫cosjxsinkxdx where k and j are Even
Evaluate ∫sin4xdx.
Solution
Since the power on sinx is even (k=4) and the power on cosx is even (j=0), we must use strategy 3. Thus,
Since cos2(2x) has an even power, substitute cos2(2x)=12+12cos(4x):
Checkpoint 3.8
Evaluate ∫cos3xdx.
Checkpoint 3.9
Evaluate ∫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(bx), cos(ax), and 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(bx), cos(ax), and cos(bx), use the substitutions
These formulas may be derived from the sum-of-angle formulas for sine and cosine.
Example 3.13
Evaluating ∫sin(ax)cos(bx)dx
Evaluate ∫sin(5x)cos(3x)dx.
Solution
Apply the identity sin(5x)cos(3x)=12sin(2x)+12sin(8x). Thus,
Checkpoint 3.10
Evaluate ∫cos(6x)cos(5x)dx.
Integrating Products and Powers of tanx and secx
Before discussing the integration of products and powers of tanx and secx, it is useful to recall the integrals involving tanx and secx we have already learned:
- ∫sec2xdx=tanx+C
- ∫secxtanxdx=secx+C
- ∫tanxdx=ln|secx|+C
- ∫secxdx=ln|secx+tanx|+C.
For most integrals of products and powers of tanx and secx, we rewrite the expression we wish to integrate as the sum or difference of integrals of the form ∫tanjxsec2xdx or ∫secjxtanxdx. As we see in the following example, we can evaluate these new integrals by using u-substitution.
Example 3.14
Evaluating ∫secjxtanxdx
Evaluate ∫sec5xtanxdx.
Solution
Start by rewriting sec5xtanx as sec4xsecxtanx.
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Checkpoint 3.11
Evaluate ∫tan5xsec2xdx.
We now take a look at the various strategies for integrating products and powers of secx and tanx.
Problem-Solving Strategy
Integrating ∫tankxsecjxdx
To integrate ∫tankxsecjxdx, use the following strategies:
- If j is even and j≥2, rewrite secjx=secj−2xsec2x and use sec2x=tan2x+1 to rewrite secj−2x in terms of tanx. Let u=tanx and du=sec2xdx.
- If k is odd and j≥1, rewrite tankxsecjx=tank−1xsecj−1xsecxtanx and use tan2x=sec2x−1 to rewrite tank−1x in terms of secx. Let u=secx and du=secxtanxdx. (Note: If j is even and k is odd, then either strategy 1 or strategy 2 may be used.)
- If k is odd where k≥3 and j=0, rewrite tankx=tank−2xtan2x=tank−2x(sec2x−1)=tank−2xsec2x−tank−2x. It may be necessary to repeat this process on the tank−2x term.
- If k is even and j is odd, then use tan2x=sec2x−1 to express tankx in terms of secx. Use integration by parts to integrate odd powers of secx.
Example 3.15
Integrating ∫tankxsecjxdx when j is Even
Evaluate ∫tan6xsec4xdx.
Solution
Since the power on secx is even, rewrite sec4x=sec2xsec2x and use sec2x=tan2x+1 to rewrite the first sec2x in terms of tanx. Thus,
Example 3.16
Integrating ∫tankxsecjxdx when k is Odd
Evaluate ∫tan5xsec3xdx.
Solution
Since the power on tanx is odd, begin by rewriting tan5xsec3x=tan4xsec2xsecxtanx. Thus,
Example 3.17
Integrating ∫tankxdx where k is Odd and k≥3
Evaluate ∫tan3xdx.
Solution
Begin by rewriting tan3x=tanxtan2x=tanx(sec2x−1)=tanxsec2x−tanx. Thus,
For the first integral, use the substitution u=tanx. For the second integral, use the formula.
Example 3.18
Integrating ∫sec3xdx
Integrate ∫sec3xdx.
Solution
This integral requires integration by parts. To begin, let u=secx and dv=sec2xdx. These choices make du=secxtanx and v=tanx. Thus,
We now have
Since the integral ∫sec3xdx has reappeared on the right-hand side, we can solve for ∫sec3xdx by adding it to both sides. In doing so, we obtain
Dividing by 2, we arrive at
Checkpoint 3.12
Evaluate ∫tan3xsec7xdx.
Reduction Formulas
Evaluating ∫secnxdx for values of n where n is odd requires integration by parts. In addition, we must also know the value of ∫secn−2xdx to evaluate ∫secnxdx. The evaluation of ∫tannxdx also requires being able to integrate ∫tann−2xdx. 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 secx or tanx with the integral of a lower power of secx or tanx.
Rule: Reduction Formulas for ∫secnxdx and ∫tannxdx
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.
Example 3.19
Revisiting ∫sec3xdx
Apply a reduction formula to evaluate ∫sec3xdx.
Solution
By applying the first reduction formula, we obtain
Example 3.20
Using a Reduction Formula
Evaluate ∫tan4xdx.
Solution
Applying the reduction formula for ∫tan4xdx we have
Checkpoint 3.13
Apply the reduction formula to ∫sec5xdx.
Section 3.2 Exercises
Fill in the blank to make a true statement.
sec2x−1=_______
Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power.
cos2x=_______
Evaluate each of the following integrals by u-substitution.
∫√cosxsinxdx
∫sin7(2x)cos(2x)dx
∫tan2xsec2xdx
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.)
∫cos3xdx
∫cos5xdx
∫sin3xcos3xdx
∫√sinxcos3xdx
∫tan(5x)dx
∫tanxsec3xdx
∫cotxdx
∫tan3x√secxdx
For the following exercises, find a general formula for the integrals.
∫sinaxcosaxdx.
Use the double-angle formulas to evaluate the following integrals.
∫π0sin4xdx
∫sin2xcos2xdx
∫sin2xcos2(2x)dx
For the following exercises, evaluate the definite integrals. Express answers in exact form whenever possible.
∫π0sin3xsin5xdx
∫π−πcos2(3x)dx
∫4π0cos(x/2)sin(x/2)dx
∫π/3−π/3√sec2x−1dx
Find the area of the region bounded by the graphs of the equations y=sinx,y=sin3x,x=0,andx=π2.
Find the area of the region bounded by the graphs of the equations y=cos2x,y=sin2x,x=−π4,andx=π4.
A particle moves in a straight line with the velocity function v(t)=sin(ωt)cos2(ωt). Find its position function x=f(t) if f(0)=0.
For the following exercises, solve the differential equations.
dydx=sin2x. The curve passes through point (0,0).
Find the length of the curve y=ln(cscx),π4≤x≤π2.
Find the volume generated by revolving the curve y=cos(3x) about the x-axis, 0≤x≤π36.
For the following exercises, use this information: The inner product of two functions f and g over [a,b] is defined by f(x)·g(x)=〈f,g〉=∫baf·gdx. Two distinct functions f and g are said to be orthogonal if 〈f,g〉=0.
Evaluate ∫π−πsin(mx)cos(nx)dx.
For each pair of integrals, determine which one is more difficult to evaluate. Explain your reasoning.
∫sin456xcosxdx or ∫sin2xcos2xdx