Math 1B: Discussion ExercisesGSI: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Find two or three classmates and a few feet of chalkboard. Introduce yourself to your newfriends, and write all of your names at the top of the chalkboard. As a group, try your hand atthe following exercises. Be sure to discuss how to solve the exercises — how you get the solutionis much more important than whether you get the solution. If as a group you agree that you allunderstand a certain type of exercise, move on to later problems. You are not expected to solve allthe exercises: some are very hard.Many of the exercises are from Single Variable Calculus: Early Transcendentals for UC Berkeleyby James Stewart; these are marked with an §. Others are my own, are from the mathematicalfolklore, or are independently marked.Here’s a hint: drawing pictures — e.g. sketching graphs of functions — will always make theproblem easier.Trigonometric IntegralsThe most important rule to know for trigonometric integrals is the Pythagorean identity:cos2x + sin2x = 1This lets us translate between (squares of) cosines and (squares of) sines. This is helpful for findingu-substitutions, since cos0= −sin and sin0= cos. For example, to integrate cos7(x), we can breakoff a cos and write the rest in terms of sin, and then substitute:Rcos7x dx =R(cos2x)3cos x dx =R(1 −sin2x)3cos x dx =R(1 −u2)3(du) =R(−u6+ 3u4−3u2+ 1)du = u7/7 −3u5/5 + u3−u + C =sin7x/7 − 3 sin5x/5 + sin3x − sin x + C. This trick turns the integral of cosnx sinmx into theintegral of a polynomial provided that n and m are non-negative and at least one of n and m areodd. (When they can be negative, but still at least one is odd, we get rational functions, which wewill learn how to integrate in section 7.4.)By dividing the Pythagorean identity by sin2or cos2, we get two more versions of the rule:1 + tan2x = sec2x and 1 + cot2x = csc2x . Since tan0= sec2and sec0= tan sec, we can integratetannsecmif m is even or n is odd. (It’s similar for cot and csc.)Sometimes, though, these aren’t enough. Then it’s important the remember the double-angleformulas:sin2x =12(1 − cos 2x) cos2x =12(1 + cos 2x) sin x cos x =12sin 2xThe product-to-sum formulas are also occasionally helpful:2 sin A sin B = cos(A − B) − cos(A + B)2 sin A cos B = sin(A − B) + sin(A + B)2 cos A cos B = cos(A − B) + cos(A + B)11. § Evaluate the integrals:(a)Zsin6x cos3x dx (b)Zπ/20cos5x dx (c)Zsin3√x√xdx(d)Zx cos2x dx (e)Zcos θ cos5(sin θ) dθ (f)Zcot5θ sin4θ dθ(g)Ztan3(2x) sec5(2x) dx (h)Ztan6(ay) dy (i)Zsin φcos3φdφ(j)Zcsc4x cot6x dx (k)Zcos x + sin xsin 2xdx (l)Zdxcos x − 1(m)Z3π/2x=0(sin x + cos x)3dx (n)Zcos 3x sin 2x dx (o)Zcos(2x + 1) cos(4x − 2) sin(x) dx2. § EvaluateRsin x cos x dx in four different ways: (a) by substituting u = cos x; (b) by substi-tuting u = sin x; (c) by using the double angle formula for sin 2x; (d) by integrating by parts.Explain the different appearances of the answers.3. Let a be a number such that 0 < a < π/2. Compute the volume obtained by rotating theregion bounded by the curvesy = tan x, y = 0, x = aabout the x-axis. Your answer should be a function of a.4. Find the average value of sin2x:12πR2π0sin2x dx. Find the average values of sin4x andsin2x cos2x.5. § Let m and n be positive integers. Prove that:(a)Zπ−πsin mx cos nx dx = 0(b)Zπ−πsin mx sin nx dx =0 if m 6= nπ if m = n(c)Zπ−πcos mx cos nx dx =0 if m 6= nπ if m = n6. (a) Use integration by parts to find a reduction formula forRπ/2x=0cosnx dx.(b) Let n = 2k + 1 be odd. Make a substitution to turnRπ/2x=0cosnx dx into a polynomialintegral. For any particular value of k you could expand this out and integrate. Instead,find a reduction formula for this integral.(c) When n = 2k + 1 is odd, solve the reduction formulas from parts (a) and (b) to find thevalue ofRπ/2x=0cosnx dx. Hint: what isRπ/2x=0cos x dx?(d) When n = 2k is even, the method in part (b) doesn’t work directly, and using double-angle formulas would be extremely messy. Solve the reduction formula from part (a) toevaluate the integral. Hint: what
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