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## 7.2 Trigonometric Integrals

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**TECHNIQUES OF INTEGRATION**7.2Trigonometric Integrals • In this section, we will learn: • How to use trigonometric identities to integrate • certain combinations of trigonometric functions.**TRIGONOMETRIC INTEGRALS**• We start with powers of sine and cosine.**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 ∫sinnx 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.**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.**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 = sec2xdx.**TANGENT & SECANT INTEGRALS**Example 5 • We have:**TANGENT & SECANT INTEGRALS**Example 6 • Find∫ tan5 θsec7θ dθ • If we separate a sec2θfactor, as in the preceding example, we are left with a sec5θfactor. • 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 θtan θdθ:**TANGENT & SECANT INTEGRALS**• The preceding examples demonstrate strategies for evaluating integrals in the form ∫tanmx secnx for two cases—which we summarize here.**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.**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.**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