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Math 142.009, Problem Set 4 Solutions (Partial) Due Wednesday, Feb. 9, 2011 From Text Calculus: Concepts and Contexts, 3rd edition, by James Stewart Section 5.6, pp. 398-399, # 1, 4, 17, 18, 21, 25, 28, 37, 39 Section 5.7, pp. 404-405, # 1, 2, 4, 5, 9, 10 Additional Problems: Evaluate the following indefinite integrals: Z 1.) x3 ln x dx We integrate by parts, which means that we apply the integration by parts formula Z Z (1) u dv = uv − v du. Because it is easier to differentiate ln x that to integrate it, we let u = ln x. Then since the integral we are R 1 1 doing must be of the form u dv, we let dv = x3 dx. Then du = dx and v = x4 . Using equation (1), x 4 we get Z Z Z 1 1 4 1 1 1 · x dx = x4 ln x − x3 dx x3 ln x dx = x4 ln x − 4 x 4 4 4 1 1 1 1 1 = x4 ln x − · x4 + C = x4 ln x − x4 + C. 4 4 4 4 16 2.) Z x sec2 x dx d tan x = sec2 x. Hence it seems reasonable to integrate by parts, letting u = x and dx dv = sec2 dx. Then du = dx and v = tan x. Hence by equation (1), we get Z Z Z sin x 2 x sec x dx = x tan x − tan x dx = x tan x − dx. cos x Z 0 d f Since cos x = − sin x, the last integral is of the form dx, so dx f Z sin x dx = ln | cos x| + C, − cos x Recall that (formally, you can do this integral by making the substitution y = cos x). Hence Z x sec2 x dx = x tan x + ln | cos x| + C. 3.) Z y 2 e−y dy We integrate by parts, letting u = y 2 and dv = e−y dy. Then du = 2y dy and v = −e−y . Hence using equation (1), Z Z Z 2 −y 2 −y −y 2 −y y e dy = −y e − −e 2y dy = −y e + 2ye−y dy. We do the last integral by integrating by parts again. Let u = 2y and dv = e−y dy. Then du = 2 dy and v = −e−y . Hence Z Z Z −y −y −y −y 2ye dy = −2ye − −2e dy = −2ye + 2 e−y dy = −2ye−y − 2e−y + C. Putting this together, we get Z 4.) Z y 2 e−y dy = −y 2 e−y − 2ye−y − 2e−y + C. sin3 x cos4 x dx Since one of the trig functions is raised to an odd power, we can reduce it to a single power by using the identity sin2 x + cos2 x = 1. Then sin2 x = 1 − cos2 x, so Z Z Z 3 4 2 4 sin x cos x dx = sin x sin x cos x dx = sin x(1 − cos2 x) cos4 x dx. Now we make the substitution u = cos x. Then du = − sin x dx. Hence Z Z Z Z 2 4 2 4 4 6 sin x(1 − cos x) cos x dx = − (1 − u )u du = − u − u du = u6 − u4 du 1 1 1 1 = u7 − u5 + C = cos7 x − cos5 x + C. 7 5 7 5 5.) Z cos4 x dx Since the trigonometric function in this integral is raised to an even power, we must use the trigonometric identity 1 1 (2) cos2 θ = + cos 2θ. 2 2 Using (2) with θ = x gives 2 Z Z Z 2 1 1 4 2 cos x dx = cos x dx = + cos 2x dx 2 2 Z 1 1 1 + 2 · cos 2x + cos2 x dx = 4 4 4 1 = 4 Z 1 + 2 cos 2x + cos2 2x dx. 1 1 The last integral includes the term cos2 2x, so we use (2) with θ = 2x to write cos2 2x = + cos 4x. 2 2 Hence Z Z 1 1 1 1 1 + 2 cos 2x + cos2 2x dx = 1 + 2 cos 2x + + cos 4x dx 4 4 2 2 Z 3 1 1 + 2 cos 2x + cos 4x dx = 4 2 2 1 1 3 x + sin 2x + sin 4x + C = 4 2 8 3 1 1 = x + sin 2x + sin 4x + C. 8 4 32 6.) Z x2 √ 1 dx 4 − x2 √ We make the trigonometric substitution x = 2 sin θ (because of the term 4 − x2 in the integral). Then dx = 2 cos θ dθ and 4 − x2 = 4 − 4 sin2 θ = 4(1 − sin2 θ) = 4 cos2 θ. √ Hence 4 cos2 θ = 2 cos θ, so Z Z 1 1 √ dx = 2 cos θ dθ 2 4 sin θ · 2 cos θ x2 4 − x2 Z Z 1 1 1 csc2 θ dθ = 2 dθ = 4 4 sin θ 1 1 cos θ = − cot θ + C == − + C. 4 4 sin θ To put the answer back in terms of x, we draw a right triangle which represents our substitution x = 2 sin θ, or sin θ = x/2, with side opposite the angle θ labelled √x and the hypoteneuse labelled 2. Then by the Pythagorean theorem the side adjacent to θ has length 4 − x2 : 2 √ θ x 4 − x2 Hence cos θ = Therefore Z √ 4 − x2 2 and sin θ = 1 1 cos θ 1 √ dx = − +C =− · 2 2 4 sin θ 4 x 4−x √ 4 − x2 =− + C. 4x √ x . 2 4 − x2 2 · +C 2 x