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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.

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**TRIGONOMETRIC INTEGRALS**

We start with powers of sine and cosine.

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**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.

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**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

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**SINE AND COSINE INTEGRALS**

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

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**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.

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**SINE AND COSINE INTEGRALS**

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

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**SINE AND COSINE INTEGRALS**

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

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**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.

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**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:

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**SINE AND COSINE INTEGRALS**

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

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**SINE AND COSINE INTEGRALS**

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

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**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.

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**SINE AND COSINE INTEGRALS**

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

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**SINE AND COSINE INTEGRALS**

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

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**SINE AND COSINE INTEGRALS**

Example 3 However, using the half-angle formula for sin2x, we have:

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**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

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**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.

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**SINE AND COSINE INTEGRALS**

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

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**SINE AND COSINE INTEGRALS**

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

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**SINE AND COSINE INTEGRALS**

Example 4 This gives:

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**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.

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**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.

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**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.

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

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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

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**TANGENT & SECANT INTEGRALS**

We can use a similar strategy to evaluate integrals of the form

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**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.

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**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.

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**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.

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**TANGENT & SECANT INTEGRALS**

Example 5 We have:

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**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.

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**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.

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**TANGENT & SECANT INTEGRALS**

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

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**TANGENT & SECANT INTEGRALS**

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

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**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.

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**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.

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**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

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**TANGENT & SECANT INTEGRALS**

We will need to be able to integrate tan x by using a substitution,

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**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.

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**TANGENT & SECANT INTEGRALS**

First, we multiply numerator and denominator by sec x + tan x:

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**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:

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**TANGENT & SECANT INTEGRALS**

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

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**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.

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**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.

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**TANGENT & SECANT INTEGRALS**

Example 8 Find Here, we integrate by parts with

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**TANGENT & SECANT INTEGRALS**

Example 8 Then,

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**TANGENT & SECANT INTEGRALS**

Example 8 Using Formula 1 and solving for the required integral, we get:

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**TANGENT & SECANT INTEGRALS**

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

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**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

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

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**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

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**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):

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