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Calculus II, Section 7.3, #8 Trigonometric Integrals Evaluate the integral.1 Z dt √ 2 t t2 − 16 Z dt √ = t2 t2 − 16 The radical has the form Z 1 √ dt t2 t2 − 4 2 √ x2 − a2 so we make the subsitution Let t = 4 sec (θ) so dt = 4 sec (θ) tan (θ) dθ Then we have t2 = 16 sec2 (θ) and t2 − 16 = 16 sec2 (θ) − 16 = 16 sec2 (θ) − 1 Now 1 + tan2 (θ) = sec2 (θ), so sec2 (θ) − 1 = tan2 (θ), and we have t2 − 16 = 16 tan2 (θ) so q p t2 − 16 = 16 tan2 (θ) p = 4 tan2 (θ) = 4 |tan (θ)| = 4 tan (θ) , θ ∈ QI or III. Since √ For |tan (θ)| to equal tan (θ), we must have tan (θ) nonnegative, which means we must have θ in Quadrant I or III. We substitute all these pieces in terms of θ into the integral Z Z 1 1 √ dt = · 4 sec (θ) tan (θ) dθ 16 sec2 (θ) · 4 tan (θ) t2 t2 − 16 Z 1 dθ = 16 sec (θ) Z 1 1 · dθ = 16 sec (θ) Z 1 · cos (θ) dθ = 16 1 = · (sin (θ) + C) 16 1 1 sin (θ) + C = 16 16 1 Stewart, Calculus, Early Transcendentals, p. 491, #8. a2 = |a|. Calculus II Trigonometric Integrals Since C is an arbitrary constant, = 1 16 C is also an arbitrary constant, so we’ll just write 1 sin (θ) + C 16 We have succeeded in integrating our function, but the result is in terms of θ rather than t, so we’ll try to substitute back to get the result in terms √ of t. If we look at our initial substitution, t = 4 sec (θ), and other calculations, t2 − 16 = 16 tan2 (θ) or t2 − 16 = 4 tan (θ), none of them involve the sin (θ) that we got from the integration. In this case, we use t = 4 sec (θ) and a right triangle to express sin (θ) in terms of t. Since t = 4 sec (θ), we have t=4· 1 cos (θ) or t 4 cos (θ) = t and we use this relationship to sketch the triangle showun in Figure 1. (We could’ve used 4t = sec (θ), but most of us are more comfortable with the cosine function.) opp. θ 4 Figure 1 From Pythagorean Theorem, we get 42 + opp.2 = t2 16 + opp.2 = t2 opp.2 = t2 − 16 And since the opposite side is a length and thus nonnegative, p opp. = t2 − 16 In Figure 2, the three sides of the triangle are labeled. From the figure, we see √ t2 − 16 sin (θ) = t √ t2 − 16 t θ 4 Figure 2 Finally, substituting gives us Z 1 1 √ dt = sin (θ) + C 16 t2 t2 − 16 √ 1 t2 − 16 = · +C 16 t √ t2 − 16 = +C 16t Thus, Z 1 √ dt = 2 2 t t − 16 √ t2 − 16 +C 16t