Integration Quiz Answers: Techniques and Calculations, Exercises of Calculus

Download Integration Quiz Answers: Techniques and Calculations and more Exercises Calculus in PDF only on Docsity! Answer Key for Quiz 3 (section A) 1. Following the hint we look at (x + 2)2 − (x2 + 4) = x2 + 4x + 4− x2 − 4 = 4x. Therefore ∫ x dx (x + 2)(x2 + 4) = 1 4 ∫ 4x dx (x + 2)(x2 + 4) = 1 4 ∫ (x + 2)2 − (x2 + 4) (x + 2)(x2 + 4) dx = 1 4 ∫ (x + 2)2 dx (x + 2)(x2 + 4) − 1 4 ∫ (x2 + 4) dx (x + 2)(x2 + 4) = 1 4 ∫ (x + 2) dx x2 + 4 − 1 4 ∫ dx x + 2 . The first integral can be looked up; it’s #25 in the table with a = 2 = c and b = 1. Or we can break it apart: ∫ x dx (x + 2)(x2 + 4) = 1 4 ∫ x dx x2 + 4 + 1 2 ∫ dx x2 + 4 − 1 4 ∫ dx x + 2 , and now we can look up the second integral and substitute u = x2 + 4 in the first, so that du = 2x dx and therefore du2 = x dx. This makes the first integral into a log, and the third is also a log, so we finally have ∫ x dx (x + 2)(x2 + 4) = 1 8 ∫ du u + 1 2 ∫ dx x2 + 4 − 1 4 ∫ dx x + 2 = 1 8 ln |u|+ 1 2 ( 1 2 arctan x 2 ) − 1 4 ln |x + 2|+ C = 1 8 ln ( x2 + 4 ) + 1 4 arctan x 2 − 1 4 ln |x + 2|+ C. 2. I like the second suggestion the best: if we let u = √ 1+t2 t then du = t 12 (1 + t 2)− 1 2 (2t)− (1 + t2) 12 t2 dt = t2√ 1+t2 −√1 + t2 t2 dt = t2 − (1 + t2) t2 √ 1 + t2 dt = −dt t2 √ 1 + t2 , so ∫ dt t2 √ 1 + t2 = ∫ (−du) = −u + C = − √ 1 + t2 t + C. The third suggestion is nice too, if you don’t mind a little trigonometry: if t = tan θ then dt = sec2 θ dθ and √ 1 + t2 = √ 1 + tan2 θ = √ sec2 θ = sec θ, so the integral becomes ∫ dt t2 √ 1 + t2 = ∫ sec2 θ dθ tan2 θ sec θ = ∫ sec θ dθ tan2 θ . This improves a lot if we multiply top and bottom by cos2 θ: ∫ sec θ dθ tan2 θ = ∫ sec θ dθ tan2 θ cos2 θ cos2 θ = ∫ cos θ dθ sin2 θ . Now let v = sin θ, so that dv = cos θ dθ and we have ∫ cos θ dθ sin2 θ = ∫ dv v2 = −1 v + C = − 1 sin θ + C.