1. if both the powers m and n are even, rewrite both trig functions using the identities in (1)

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1 Section 7. Avance Integration Techniques: Trigonometric Integrals We will use the following ientities quite often in this section; you woul o well to memorize them. sin x 1 cos(x cos x 1+cos(x (1 cos(x 1 sin x cos(x cos x 1 sec x 1 + tan x csc x 1 + cot x When attempting to integrate a function built up from trigonometric functions, there are often many ifferent possibilities for choosing an integration technique. For example, we can solve sin x cos x using the u-substitution u cos x. The same substitution coul be use to fin tan x if we note that tan x sin x cos x. We can use integration by parts to solve sin(5x cos(3x. However, there are many other trigonometric functions whose integrals can not be evaluate so easily. In this section, we will look at multiple techniques for hanling integrals of several ifferent types of trig functions. Integrals of the form sin m x cos n x To integrate a function of the form sin m x cos n x, we will use one of the two following methos: 1. if both the powers m an n are even, rewrite both trig functions using the ientities in (1. Otherwise, we will rewrite the function so that only one power of sin x (or one power of cos x appears; this will allow us to make a helpful substitution: (a If m k + 1 is o, then rewrite sin m x sin k+1 x (sin x(sin k x (sin x(sin x k (sin x(1 cos x k, then use the u-substitution u cos x. (b If n k + 1 is o, then rewrite cos n x cos k+1 x (cos x(cos k x (cos x(cos x k (cos x(1 sin x k, then use the u-substitution u sin x. Examples: Fin cos 3 (x. Since cos(x has an o power, let s rewrite cos 3 (x cos(x cos (x cos(x(1 sin (x. Then cos 3 (x cos(x(1 sin (x. 1

2 Section 7. We will nee the substitution u sin(x so that u cos(x. Now we can finish the problem: cos 3 (x cos(x(1 sin (x 1 1 u u using the substitution u sin(x (u 1 13 u3 + C 1 u 1 6 u3 + C 1 sin(x 1 6 sin3 (x + C. Fin sin 3 x cos 5 x. Since both trig functions have o powers, we will rewrite one of them using the Pythagorean ientity. Let s try sin 3 x cos 5 x sin 3 x cos x cos x sin 3 x(cos x cos x sin 3 x(1 sin x cos x. As in the previous example, we can use a simple u-substitution to finish the problem. u sin x so that u cos x. Then sin 3 x(1 sin x cos x u 3 (1 u u u 3 (1 u + u u u 3 u 5 + u 7 u Set 1 u 6 u6 + 1 u + C 1 sin x 1 3 sin6 x + 1 sin x + C. Fin cos (x. Since there are no o powers in this function, we will rewrite cos (x 1+cos(x using the

3 Section 7. equation in (1. Then the integral calculation is fairly routine: 1 + cos(x cos (x cos(x 1 (x + 1 sin(x + C using the substitution u x 1 x + 1 sin(x + C. an Evaluate cos x sin x. Since both the powers of cos x an sin x are even, we will write cos x 1 + cos(x sin x (sin x ( 1 cos(x. Then ( ( cos x sin 1 + cos(x 1 cos(x x ( ( 1 + cos(x 1 cos(x + cos (x 1 1 cos(x + cos (x + cos(x cos (x + cos 3 (x 1 1 cos(x cos (x + cos 3 (x (x 1 1 sin x cos (x + cos 3 (x. We have alreay showe that cos (x 1 x + 1 sin(x + C an cos 3 (x 1 sin(x 1 6 sin3 (x + C, so finally we have ( cos x sin x 1 x 1 sin x 1 x 1 sin(x + 1 sin(x 1 6 sin3 (x + C. 3

4 Section 7. Integrating powers of tan x, sec x, csc x, an cot x To integrate powers of the other trig functions, we will often nee to use u-substitution or integration by parts together with the pythagorean ientities; if possible, we will nee to take avantage of the fact that tan x sec x, sec x sec x tan x, Example: Evaluate csc x. csc x csc x cot x, an cot x csc x. Writing csc x (csc x(csc x (1 + cot x(csc x is avantageous, as it will allow us to use the substitution u cot x: csc x (1 + cot x(csc x (1 + u u using u cot x an u csc x u 1 3 u3 + C cot x 1 3 cot3 x + C. Eliminating square roots If the function we wish to integrate involves the square root of some trigonometric function, we may be able to eliminate the root by using the pythagorean ientities or the ientities from (1. Examples: Evaluate cos y + 1y The ientity cos x 1+cos(x can help us here. Setting y x, so that x y, the ientity becomes cos ( y 1+cos y. We woul like to replace the quantity cos y +1; solving for this quantity in the above ientity, we have cos y + 1 cos ( y. So we may rewrite the integral as cos(y ( y + 1 cos y ( y cos y ( y sin + C. Fin csc θ 1θ.

5 Section 7. Since csc θ 1 cot θ, let s rewrite csc cot θ 1θ θθ cot θθ. Since cot θ cos θ, we can integrate the function using a substitution; setting u sin θ so that sin θ u cos θθ, we have csc θ 1θ cot θθ cos θ sin θ θ 1 u u ln u + C ln(sin θ + C. A brief asie We have not yet learne how to evaluate sec x, an as we will nee to know this integral in future sections, let s go ahea an compute it. It turns out that the best way to evaluate the integral is by using Mathemagic: note that sec x can be rewritten as sec x sec x sec x + sec x tan x. This may seem pointless, but it will actually allow us to use a basic substitution to evaluate the integral. Setting u so that u sec x tan x + sec x, we have sec x + sec x tan x sec x 1 u u ln u + C ln + C. 5