# 7.2 Trigonometric Integrals

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7.2 Trigonometric Integrals
TECHNIQUES OF INTEGRATION 7.2 Trigonometric Integrals In this section, we will learn: How to use trigonometric identities to integrate certain combinations of trigonometric functions.

TRIGONOMETRIC INTEGRALS

SINE AND COSINE INTEGRALS
Example 1 Evaluate ∫ cos3x dx Simply substituting u = cos x is not helpful, since then du = – sin x dx. In order to integrate powers of cosine, we would need an extra sin x factor. Similarly, a power of sine would require an extra cos x factor.

SINE AND COSINE INTEGRALS
Example 1 Thus, here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine using the identity sin2x + cos2x = 1: cos3x = cos2x . cosx = (1 - sin2x) cosx

SINE AND COSINE INTEGRALS
Example 1 We can then evaluate the integral by substituting u = sin x. So, du = cos x dx and

SINE AND COSINE INTEGRALS
In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor. The remainder of the expression can be in terms of cosine.

SINE AND COSINE INTEGRALS
We could also try only one cosine factor. The remainder of the expression can be in terms of sine.

SINE AND COSINE INTEGRALS
The identity sin2x + cos2x = 1 enables us to convert back and forth between even powers of sine and cosine.

SINE AND COSINE INTEGRALS
Example 2 Find ∫ sin5x cos2x dx We could convert cos2x to 1 – sin2x. However, we would be left with an expression in terms of sin x with no extra cos x factor.

SINE AND COSINE INTEGRALS
Example 2 Instead, we separate a single sine factor and rewrite the remaining sin4x factor in terms of cos x. So, we have:

SINE AND COSINE INTEGRALS
Example 2 Substituting u = cos x, we have du =-sin x dx. So,

SINE AND COSINE INTEGRALS
The figure shows the graphs of the integrand sin5x cos2x in Example 2 and its indefinite integral (with C = 0).

SINE AND COSINE INTEGRALS
In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. If the integrand contains even powers of both sine and cosine, this strategy fails.

SINE AND COSINE INTEGRALS
In that case, we can take advantage of the following half-angle identities:

SINE AND COSINE INTEGRALS
Example 3 Evaluate If we write sin2x = 1 - cos2x, the integral is no simpler to evaluate.

SINE AND COSINE INTEGRALS
Example 3 However, using the half-angle formula for sin2x, we have:

SINE AND COSINE INTEGRALS
Example 3 Notice that we mentally made the substitution u = 2x when integrating cos 2x. Another method for evaluating this integral was given in Exercise 43 in Section 7.1

SINE AND COSINE INTEGRALS
Example 4 Find We could evaluate this integral using the reduction formula for ∫ sinn x dx (Equation 7 in Section 7.1) together with Example 3.

SINE AND COSINE INTEGRALS
Example 4 However, a better method is to write and use a half-angle formula:

SINE AND COSINE INTEGRALS
Example 4 As cos2 2x occurs, we must use another half-angle formula:

SINE AND COSINE INTEGRALS
Example 4 This gives:

SINE AND COSINE INTEGRALS
To summarize, we list guidelines to follow when evaluating integrals of the form where m ≥ 0 and n ≥ 0 are integers.

If the power of cosine is odd (n = 2k + 1), save one cosine factor.
STRATEGY A If the power of cosine is odd (n = 2k + 1), save one cosine factor. Use cos2x = 1 - sin2x to express the remaining factors in terms of sine: Then, substitute u = sin x.

If the power of sine is odd (m = 2k + 1), save one sine factor.
STRATEGY B If the power of sine is odd (m = 2k + 1), save one sine factor. Use sin2x = 1 - cos2x to express the remaining factors in terms of cosine: Then, substitute u = cos x.

STRATEGIES Note that, if the powers of both sine and cosine are odd, either (A) or (B) can be used.

STRATEGY C If the powers of both sine and cosine are even, use the half-angle identities Sometimes, it is helpful to use the identity

TANGENT & SECANT INTEGRALS
We can use a similar strategy to evaluate integrals of the form

TANGENT & SECANT INTEGRALS
As (d/dx)tan x = sec2x, we can separate a sec2x factor. Then, we convert the remaining (even) power of secant to an expression involving tangent using the identity sec2x = 1 + tan2x.

TANGENT & SECANT INTEGRALS
Alternately, as (d/dx) sec x = sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant.

TANGENT & SECANT INTEGRALS
Example 5 Evaluate ∫ tan6x sec4x dx If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x = 1 + tan2x. Then, we can evaluate the integral by substituting u = tan x so that du = sec2x dx.

TANGENT & SECANT INTEGRALS
Example 5 We have:

TANGENT & SECANT INTEGRALS
Example 6 Find ∫ tan5 θ sec7θ d θ If we separate a sec2θ factor, as in the preceding example, we are left with a sec5θ factor. This is not easily converted to tangent.

TANGENT & SECANT INTEGRALS
Example 6 However, if we separate a sec θ tan θ factor, we can convert the remaining power of tangent to an expression involving only secant. We can use the identity tan2θ = sec2θ – 1.

TANGENT & SECANT INTEGRALS
Example 6 We then evaluate the integral by substituting u = sec θ, so du = sec θ tan θ dθ:

TANGENT & SECANT INTEGRALS
The preceding examples demonstrate strategies for evaluating integrals in the form ∫ tanmx secnx for two cases—which we summarize here.

If the power of secant is even (n = 2k, k ≥ 2) save sec2x.
STRATEGY A If the power of secant is even (n = 2k, k ≥ 2) save sec2x. Then, use tan2x = 1 + sec2x to express the remaining factors in terms of tan x: Then, substitute u = tan x.

If the power of tangent is odd (m = 2k + 1), save sec x tan x.
STRATEGY B If the power of tangent is odd (m = 2k + 1), save sec x tan x. Then, use tan2x = sec2x – 1 to express the remaining factors in terms of sec x: Then, substitute u = sec x.

For other cases, the guidelines are not as clear-cut.
OTHER INTEGRALS For other cases, the guidelines are not as clear-cut. We may need to use: Identities Integration by parts A little ingenuity

TANGENT & SECANT INTEGRALS
We will need to be able to integrate tan x by using a substitution,

TANGENT & SECANT INTEGRALS
Formula 1 We will also need the indefinite integral of secant: We could verify Formula 1 by differentiating the right side, or as follows.

TANGENT & SECANT INTEGRALS
First, we multiply numerator and denominator by sec x + tan x:

TANGENT & SECANT INTEGRALS
If we substitute u = sec x + tan x, then du = (sec x tan x + sec2x) dx. The integral becomes: ∫ (1/u) du = ln |u| + C Thus, we have:

TANGENT & SECANT INTEGRALS
Example 7 Find Here, only tan x occurs. So, we rewrite a tan2x factor in terms of sec2x.

TANGENT & SECANT INTEGRALS
Example 7 Hence, we use tan2x - sec2x = 1. In the first integral, we mentally substituted u = tan x so that du = sec2x dx.

TANGENT & SECANT INTEGRALS
If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. Powers of sec x may require integration by parts, as shown in the following example.

TANGENT & SECANT INTEGRALS
Example 8 Find Here, we integrate by parts with

TANGENT & SECANT INTEGRALS
Example 8 Then,

TANGENT & SECANT INTEGRALS
Example 8 Using Formula 1 and solving for the required integral, we get:

TANGENT & SECANT INTEGRALS
Integrals such as the one in the example may seem very special. However, they occur frequently in applications of integration.

COTANGENT & COSECANT INTEGRALS
Integrals of the form ∫ cotmx cscnx dx can be found by similar methods. We have to make use of the identity 1 + cot2x = csc2x

OTHER INTEGRALS Finally, we can make use of another set of trigonometric identities, as follows.

In order to evaluate the integral, use the corresponding identity.
OTHER INTEGRALS Equation 2 In order to evaluate the integral, use the corresponding identity. Integral Identity a ∫ sin mx cos nx dx b ∫ sin mx sin nx dx c ∫ cos mx cos nx dx

TRIGONOMETRIC INTEGRALS
Example 9 Evaluate ∫ sin 4x cos 5x dx This could be evaluated using integration by parts. It is easier to use the identity in Equation 2(a):