7 TECHNIQUES OF INTEGRATION TECHNIQUES OF INTEGRATION 7

7 TECHNIQUES OF INTEGRATION

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 We start with powers of sine and cosine.

SINE & COSINE INTEGRALS Example 1 Evaluate ∫ cos 3 x dx § Simply substituting u = cos x isn’t 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 & COSINE INTEGRALS Example 1 Thus, here we can separate one cosine factor and convert the remaining cos 2 x factor to an expression involving sine using the identity sin 2 x + cos 2 x = 1: cos 3 x = cos 2 x. cosx = (1 - sin 2 x) cosx

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

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

SINE & COSINE INTEGRALS The identity sin 2 x + cos 2 x = 1 enables us to convert back and forth between even powers of sine and cosine.

SINE & COSINE INTEGRALS Example 2 Find ∫ sin 5 x cos 2 x dx § We could convert cos 2 x to 1 – sin 2 x. § However, we would be left with an expression in terms of sin x with no extra cos x factor.

SINE & COSINE INTEGRALS Example 2 Instead, we separate a single sine factor and rewrite the remaining sin 4 x factor in terms of cos x. So, we have:

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

SINE & COSINE INTEGRALS The figure shows the graphs of the integrand sin 5 x cos 2 x in Example 2 and its indefinite integral (with C = 0).

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

SINE & COSINE INTEGRALS Example 3 Evaluate § If we write sin 2 x = 1 - cos 2 x, the integral is no simpler to evaluate.

SINE & COSINE INTEGRALS Example 3 However, using the half-angle formula for sin 2 x, we have:

SINE & COSINE INTEGRALS Example 3 Notice that we mentally made the substitution u = 2 x when integrating cos 2 x. § Another method for evaluating this integral was given in Exercise 43 in Section 7. 1

SINE & COSINE INTEGRALS Example 4 Find ∫ sin 4 x dx § We could evaluate this integral using the reduction formula for ∫ sinnx dx (Equation 7 in Section 7. 1) together with Example 3.

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

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

SINE & COSINE INTEGRALS This gives: Example 4

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

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

STRATEGY B If the power of sine is odd (m = 2 k + 1), save one sine factor. § Use sin 2 x = 1 - cos 2 x 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 ∫ tanmx secnx dx

TANGENT & SECANT INTEGRALS As (d/dx)tan x = sec 2 x, we can separate a sec 2 x factor. § Then, we convert the remaining (even) power of secant to an expression involving tangent using the identity sec 2 x = 1 + tan 2 x.

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 ∫ tan 6 x sec 4 x dx § If we separate one sec 2 x factor, we can express the remaining sec 2 x factor in terms of tangent using the identity sec 2 x = 1 + tan 2 x. § Then, we can evaluate the integral by substituting u = tan x so that du = sec 2 x dx.

TANGENT & SECANT INTEGRALS Example 5 We have:

TANGENT & SECANT INTEGRALS Example 6 Find ∫ tan 5 θ sec 7θ § If we separate a sec 2θ factor, as in the preceding example, we are left with a sec 5θ factor. § This isn’t 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 tan 2θ = sec 2θ – 1.

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

TANGENT & SECANT INTEGRALS The preceding examples demonstrategies 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 = 2 k, k ≥ 2) save sec 2 x. § Then, use tan 2 x = 1 + sec 2 x 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 = 2 k + 1), save sec x tan x. § Then, use tan 2 x = sec 2 x – 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 Formula 5 from Section 5. 5 :

TANGENT & SECANT INTEGRALS Formula 1 We will also need the indefinite integral of secant:

TANGENT & SECANT INTEGRALS 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 + sec 2 x). § The integral becomes: ∫ (1/u) du = ln |u| + C

TANGENT & SECANT INTEGRALS Thus, we have:

TANGENT & SECANT INTEGRALS Example 7 Find ∫ tan 3 x dx § Here, only tan x occurs. § So, we rewrite a tan 2 x factor in terms of sec 2 x.

TANGENT & SECANT INTEGRALS Example 7 Hence, we use tan 2 x - sec 2 x = 1. § In the first integral, we mentally substituted u = tan x so that du = sec 2 x 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 ∫ sec 3 x dx § 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. § We will see this in Chapter 8.

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 + cot 2 x = csc 2 x

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

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

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