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

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We start with powers of sine and cosine. TRIGONOMETRIC INTEGRALS

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SINE AND COSINE INTEGRALS Evaluate ∫ cos 3 x 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. Example 1

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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 Example 1 SINE AND COSINE INTEGRALS

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

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

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

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SINE AND COSINE INTEGRALS 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. Example 2

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

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

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SINE AND 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).

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

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

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SINE AND COSINE INTEGRALS 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 Example 3

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

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

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

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SINE AND COSINE INTEGRALS This gives: Example 4

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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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STRATEGY A If the power of cosine is odd (n = 2k + 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.

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If the power of sine is odd (m = 2k + 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. STRATEGY B

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

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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 = 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.

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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 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. Example 5

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TANGENT & SECANT INTEGRALS We have: Example 5

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TANGENT & SECANT INTEGRALS Find ∫ tan 5 θ sec 7 θ d θ If we separate a sec 2 θ factor, as in the preceding example, we are left with a sec 5 θ factor. This is not easily converted to tangent. Example 6

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TANGENT & SECANT INTEGRALS 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 2 θ = sec 2 θ – 1. Example 6

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

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TANGENT & SECANT INTEGRALS The preceding examples demonstrate strategies for evaluating integrals in the form ∫ tan m x sec n x for two cases—which we summarize here.

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If the power of secant is even (n = 2k, 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 A

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If the power of tangent is odd (m = 2k + 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. STRATEGY B

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

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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 + sec 2 x) dx. The integral becomes: ∫ (1/u) du = ln |u| + C Thus, we have:

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

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TANGENT & SECANT INTEGRALS 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. Example 7

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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 Find Here, we integrate by parts with Example 8

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TANGENT & SECANT INTEGRALS Then, Example 8

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

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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 ∫ cot m x csc n x dx can be found by similar methods. We have to make use of the identity 1 + cot 2 x = csc 2 x

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

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

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TRIGONOMETRIC INTEGRALS 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): Example 9

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