6.2 Trigonometric Integrals and Substitutions

Transcription

1 Arkansas Tech University MATH 9: Calculus II Dr. Marcel B. Finan 6. Trigonometric Integrals and Substitutions In this section, we discuss integrals with trigonometric integrands and integrals that can be transformed to trigonometric integrals by substitution. Trigonometric Integrals In order to understand the following discussion, the reader is encouraged to review trigonometric identities before proceeding with the discussion. Integrals of the form sin n xdx or cos n xdx Example 6.. Find sin 3 xdx. First, we notice that sin 3 x = sin x sin x = sin x( cos x. Using the substitution u = cos x, where du = sin xdx, we find sin 3 xdx = Example 6.. Find cos xdx. Using the trigonometric identity ( u du = u + u3 3 + C = cos x + cos3 x + C 3 cos x = + cos x

2 we find ( + cos x cos xdx = [cos x] dx = dx = ( + cos x + cos xdx = [ ( + cos x dx + cos xdx + = ( sin x x + + x + sin x = ( 3 sin x x + sin x + 8 ] dx Example 6..3 Find sin xdx. Using the trigonometric identity we find sin x = cos x dx ( cos x sin xdx = = ( cos xdx = ( sin x x + C Integrals of the form sin n x cos m xdx Example 6.. Find sin 3 x cos xdx.

3 Let u = cos x then du = sin xdx. Thus, sin 3 x cos xdx = ( cos x sin x cos xdx = ( u u du = ( u + u 6 du = u5 5 + u7 7 + C Example 6..5 Find sin x cos xdx. = cos5 x 5 + cos7 x 7 + C Using the trigonometric identity cos x + sin x =, we have sin x cos xdx = sin x( sin xdx = sin xdx sin xdx ( ( x sin x = sin3 x cos x + 3 = x = x sin x sin x Integrals of Tangent and Secant Example 6..6 Find tan xdx. + sin3 x cos x 3 ( x sin x sin xdx + C + sin3 x cos x sin x 3 8 x + C = 6 sin x + sin3 x cos x + 8 x + C 3

4 Using the substitution u = cos x, we find tan xdx = Example 6..7 Find sec xdx. First, note that sin x cos x dx = du u = ln u + C = ln cos x + C = ln sec x + C (sec x + tan x = sec x + sec x tan x = sec x(sec x + tan x. Using the substitution u = sec x + tan x, we find sec x + tan x sec xdx = sec x sec x + tan x dx sec x + sec x tan x = dx sec x + tan x du = = ln u + C u = ln sec x + tan x + C

5 Trigonometric Substitutions This subsection deals with integrands involving terms like x a, x + a, and a x. Integrands involving a x, a x a, a > 0. For each x in the interval [ a, a] there is a θ in the interval [ π, π ] such that x = a sin θ (notice that x a and recall the graph of sin x. Thus, using the substitution x = a sin θ, π θ π to obtain a x = a ( sin θ = a cos θ = a cos θ =a cos θ where we have used the Pythagorean identity cos θ +sin θ =. Notice that cos θ = cos θ since cos θ 0 in π θ π. It is important to point out here that by constructing a right triangle with one of the angle being θ then the hypotenuse of the triangle has length a, the opposite side has length x and the adjacent side has length a x. It follows that cos θ = See Figure 6... a x a. Example 6..8 Find x dx. Let x = sin θ, π < θ < π. Then Figure 6.. x = sin θ = cos θ = cos θ. Moreover, dx = cos θdθ. It follows that cos θ ( x dx = x cos θ dθ = θ + C = arcsin + C 5

6 Integrands involving a + x, a > 0. In this case, we let x = a tan θ with π < θ < π. Such a substitution leads to a + x = a + a tan θ = a ( + tan θ = a sec θ = a sec θ since + tan θ = sec θ and sec θ > 0 for π < θ < π. Remark 6.. Letting θ be the angle of a right triangle with opposite side x, adjacent side a, and hypotenuse a + x we find sec θ = a +x a. See Figure 6... Example 6..9 Find x +9 dx. Figure 6.. Let x = 3 tan θ with π < θ < π. Then x + 9 = tan θ = Moreover, dx = 3 sec θdθ. Thus, 3 sec x + 9 dx = θ 3 sec θ dθ = 9( + tan θ = 3 sec θ. sec θdθ = ln sec θ + tan θ + C. Now, considering a triangle with acute angle θ, opposite side x, and adjacent side 3 we see that sec θ = 9+x 3 and tan θ = x 3. Thus, 9 + x x + 9 dx = ln + x C Integrands Involving x a, x a or x a, a 0. Here, we let x = a sec θ with 0 θ < π or π θ < 3π so that x a = a (sec θ = a tan θ = a tan θ = a tan θ. 6

7 Example 6..0 Evaluate x 9 dx. Let x = 3 sec θ with 0 < θ < π or π < θ < 3π. Then dx = 3 sec θ tan θdθ. It follows that 3 sec θ tan θ x 9 dx = dθ 3 tan θ = sec θdθ = ln sec θ + tan θ + C x x = ln C Example 6.. Find x x dx. Let x = sec θ, 0 θ < π or π θ < 3π. Then dx = sec θ tan θdθ and x = sec θ = tan θ. Thus, sec θ tan θ x x dx = sec θ tan θ dθ = dθ = θ + C = sec x + C We can summarize the above substitutions in the following table expression substitution identity a u u = a sin θ, π θ π sin θ = cos θ a + u u = a tan θ, π < θ < π + tan θ = sec θ u a u = a sec θ, 0 θ < π, π θ < 3π sec θ = tan θ Example 6.. Find the area of the circular sector with radius r and central angle θ as 7

8 shown in Figure 6..3 Figure 6..3 The total area is the sum of the area of the triangle POQ and the area of the region PQR in the figure. We have Area of triangle POQ = PQ OQ = r cos θ sin θ and using the substitution x = r cos w we find Area of region PQR = Hence, the total area is r r cos θ r x dx 0 = r sin wdw =r θ 0 θ sin wdw = r [w cos w sin w]θ 0 = r (θ cos θ sin θ. r cos θ sin θ + r (θ cos θ sin θ = r θ 8