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MAT 136F Tutorial 4 – Integration methods II October 26–30, 2015 1. Evaluate the following integrals. Z −3x + 17 (a) dx x2 − 2x − 3 Z 4 − 4x (b) dx 2 x − 2x − 3 Hint: This is easier than the previous one. Z x2 √ (c) dx 9 − x2 Z x √ (d) dx 9 − x2 Hint: This is easier than the previous one. Z x √ dx 5 − 4x − x2 Z 3x2 + 2x + 1 dx x3 + x2 + x Z 4x4 − x2 + 6x − 1 dx 4x2 − 1 (e) (f) (g) Z (h) ex Z (i) √ dx dx 4 + e2x dx √ 4x2 + 2x 2. The integral of sec x. In your textbook, the integral of sec x was calculated using some trickery. Here we will study another method for integrating sec x, that is longer, but less dependent on a memorized trick. (a) Convince yourself that sec x = cos x . 1 − sin2 x Z (b) Try a substitution to transform the integral cos x dx into the integral of a 1 − sin2 x rational function. (c) Use partial fractions to evaluate the integral you obtained in part (b). (d) Write your answer in terms of the original variable, x. Z 1 sin x + 1 Your final answer should look like sec x dx = ln +C. 2 sin x − 1 Notice that this looks different from the answer in your textbook. However, with a little bit of work, it is possible to see that the answer that you got is really the same. 1 sin x + 1 (e*) (optional) Show that ln = ln sec x + tan x 2 sin x − 1 Hint: Start with the left-hand side, multiply √ Next, remember that 21 ln a = ln a. sin x+1 sin x−1 by sin x+1 . sin x+1 MAT 136F Tutorial 4 – Integration methods II October 26–30, 2015 Solutions −5 2 −3x + 17 = + 1. (a) Use partial fractions to get: 2 x − 2x − 3 x+1 x−3 Z −3x + 17 Answer: dx = −5 ln |x + 1| + 2 ln |x − 3| + C x2 − 2x − 3 (b) Substitution u = x2 − 2x − 3 Z 4 − 4x Answer: dx = −2 ln |x2 − 2x − 3| + C 2 x − 2x − 3 (c) Trigonometric Substitution: x = 3 sin θ Z x2 9 x x√ √ Answer: dx = sin−1 − 9 − x2 + C 2 2 3 2 9−x (d) Substitution: u = 9 − x2 Z x √ Answer: dx = −(9 − x2 )1/2 + C 2 9−x x x . Then use the trigono=p 5 − 4x − x2 9 − (x + 2)2 metric substitution x = 3 sin θ − 2. Z √ x x+2 −1 2 √ Answer: dx = − 5 − 4x − x − 2 sin +C 3 5 − 4x − x2 (e) First complete the square: √ 3x2 + 2x + 1 1 2x + 1 (f) Use partial fraction to obtain: 3 = + 2 . 2 x +x +x x x +x+1 Z 3x2 + 2x + 1 Answer: dx = ln |x| + ln |x2 + x + 1| + C x3 + x2 + x 4x4 − x2 + 6x − 1 1 2 (g) Use long division and partial fractions to get: = x2 + + . 2 4x − 1 2x − 1 2x + 1 Z 4x4 − x2 + 6x − 1 x3 1 dx = + ln |2x − 1| + ln |2x + 1| + C Answer: 4x2 − 1 3 2 (h) Substitution ex = 2 tan θ. Or, if you prefer, you can do it in two steps: first ex = u, then do a trig substitution. √ Z dx 4 + e2x √ Answer: =− +C 4ex ex 4 + e2x 1 1 (i) Complete the square. Then trig substitution 2x + = sec θ, which is the same as 2 2 sec θ = 4x + 1. Z i √ dx 1 h √ Answer: = ln 4x + 1 + 2 4x2 + 2x + C 2 4x2 + 2x 1 , multiplying by cos x to both the numerator and denominator, 2. (a) Because sec x = cos x we obtain that cos x cos x = sec x = . 2 cos x 1 − sin2 x (b) Let u = sin x. Then Z cos x dx = 1 − sin2 x Z 1 du. 1 − u2 (c) Z 1 du = 1 − u2 Z 1 2 1 1 + 1+u 1−u du = 1 (ln |1 + u| − ln |1 − u|) + C. 2 (d) Since u = sin x, 1 1 sin x + 1 (ln |1 + u| − ln |1 − u|) + C = ln + C. 2 2 sin x − 1 Now consider sin x + 1 (sin x + 1)2 = sin x − 1 (sin x − 1)(sin x + 1) . And by applying the trigonometric identity sin2 x + cos2 x = 1, what can you obtain? MAT 136F Tutorial 4 – Integration methods II October 26–30, 2015 Detailed solutions for selected questions 1. (a) Notice that the denominator x2 − 2x − 3 = (x + 1)(x − 3). We may assume the integrand is the sum of the partial fractions: A B −3x + 17 = + , 2 x − 2x − 3 x+1 x−3 where A and B are constants to be determined. Multiplying by the least common denominator, we get A(x − 3) + B(x + 1) (A + B)x − 3A + B −3x + 17 = = . 2 2 x − 2x − 3 x − 2x − 3 x2 − 2x − 3 Comparing the numerators, we obtain the following system of equations for A and B: A+B = −3 −3A + B = 17 By solving the system, we get A = −5 and B = 2, and therefore, Z Z −5 2 −3x + 17 dx = + dx x2 − 2x − 3 x+1 x−3 = −5 ln |x + 1| + 2 ln |x − 3| + C (c) Consider the trigonometric substitution: x = 3 sin θ. Then dx = 3 cos θdθ. And Z Z 9 sin2 θ x2 √ p dx = · 3 cos θ dθ 9 − x2 9 − 9 sin2 θ Z Z 9 sin2 θ cos θ = dθ = 9 sin2 θ dθ cos θ Z 1 =9 (1 − cos 2θ) dθ 2 9θ 9 sin 2θ = − + C. 2 4 Now notice that because x = 3 sin θ, we have θ = sin−1 x3 . Then x sin 2θ = 2 sin θ cos θ = 2 · · 3 r 1− x2 2 √ = x 9 − x2 . 9 9 Therefore, we obtain that Z x x√ x2 9 √ dx = sin−1 − 9 − x2 + C. 2 2 3 2 9−x (e) We first complete the square for the polynomial under the square root: 5 − 4x − x2 = 5 − (x2 + 2 · 2x + 4) + 4 = 9 − (x + 2)2 . This suggests we make a trigonometric substitution x + 2 = 3 sin θ. Then x = 3 sin θ − 2 and dx = 3 cos θdθ. So our integral becomes Z Z Z x 3 sin θ − 2 x √ p p dx = dx = · 3 cos θ dθ 5 − 4x − x2 9 − (x + 2)2 9 − 9 sin2 θ Z = (3 sin θ − 2) dθ = −3 cos θ − 2θ + C. p and cos θ = 1 − sin2 θ = Now x + 2 = 3 sin θ implies that θ = sin−1 x+2 3 q 2 1 − x+2 . After simplifying the expression, we obtain that 3 Z √ x + 2 x −1 √ dx = − 5 − 4x − x2 − 2 sin + C. 3 5 − 4x − x2 (h) We first substitute ex = u. Then du = ex dx, or dx = 1 1 du = du. x e u The integral then becomes Z Z Z 1 dx 1 1 √ √ √ = · du = du. x 2x 2 2 e 4+e u 4+u u u 4 + u2 Now consider a substitution u = 2 tan θ. Then du = 2 sec2 θdθ, so Z Z 1 1 √ du = · 2 sec2 θ dθ 2 2 2 4 tan θ · 2 sec θ u 4+u Z Z 1 sec θ 1 cos θ = dθ = dθ 2 4 tan θ 4 sin2 θ 1 =− + C. 4 sin θ Because tan θ = u/2, u ex sin θ = √ =√ . u2 + 22 4 + e2x Therefore, √ Z dx 4 + e2x √ =− + C. 4ex ex 4 + e2x