HOMEWORK SOLUTIONS MATH 1910 Sections 8.2, 8.3, 8.5 Fall 2016
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1 HOMEWORK SOLUTIONS MATH 191 Sections 8., 8., 8.5 Fall 16 Problem Evaluate using methods similar to those that apply to integral tan m xsec n x. cot x SOLUTION. Using the reduction formula for cot m x, we get cot x = 1 cot x cotx = 1 cot x ln sinx + C Problem cos (πθ)sin (πθ)dθ SOLUTION. Use the substitution u = πθ, du = πdθ, and the identity cos u = 1 sin u to write cos (πθ)sin (πθ)dθ = 1 cos usin udu = 1 (1 sin u)sin ucosudu π π Now use the substitution w = sinu, dw = cosudu: cos (πθ)sin (πθ)dθ = 1 (1 w )w dw = 1 π 5π w5 1 7π w7 +C = 1 5π sin5 (πθ) 1 7π sin7 (πθ)+c Problem Find the volume of the solid obtained by revolving y = sinx for x π about the x-axis. SOLUTION. π π(sinx) = π( x sinx ) π = π
2 Problem Evaluate the integral J = sin m x cos n x where m and n are both even using the identities sin x = 1 (1 cos x) and cos x = 1 (1 + cos x): J = cos x SOLUTION. Using the second identity, J = 1 (1+cos x) = 1 (1+ cos x+cos x) = cos x (1+cos x) Using the substitution, we see J = 1 x + 1 sin x x + 1 sin x + C = 8 x + 1 sin x + 1 sin x + C Problem I = x + 9 (a) Show that the substitution x = tanθ transforms I into 1 (b) Evaluate I in terms of θ. (c) Express I in terms of x. secθdθ SOLUTION. (a) if x = tanθ. Then = sec θdθ and x + 9 = 9tan θ + 9 = sec θ = secθ Thus, I = x + 9 = sec θdθ = 1 secθdθ secθ (b) I = 1 secθdθ = 1 ln secθ + tanθ + C (c) Since x = tanθ, we construct a right angle triangle with tanθ = x
3 From this we see that secθ = 1 x + 9, and so I = 1 ln secθ + tanθ + C = 1 ln 1 x x + C = 1 ln x x + C = 1 ln x x 1 ln + C = 1 ln x x + C Problem Evaluate using trigonometric substitution. Refer to the table of trigonometric integrals as necessary. x 9 x SOLUTION. Let x = sinθ. Then = cosθdθ and 9 x = 9 9sin θ = 9(1 sin θ) = 9cos θ x 9sin I = = θ(cosθdθ) = 9 sin θdθ = 9( θ 9 x cosθ 1 sinθcosθ) + C Since x = sinθ, we construct a right angle triangle with sinθ = x From this we see that cosθ = 9 x, and so I = 9 sin 1 ( x ) 9 9 x (x )( ) + C = 9 sin 1 ( x ) 1 x 9 x + C Problem x + x + 1 SOLUTION. First complete the squarre: x + x + 1 = x + x = (x + ) + 9
4 Let u = x +. Then = du, and I = x + x + 1 = (x + ) + 9 = du u + 9 Now let u = tanθ. Then du = sec θdθ,, u + 9 = 9tan θ + 9 = 9(tan θ + 1) = 9sec θ and I = sec θdθ secθ = secθdθ = ln secθ + tanθ + C Since u = tanθ, we construct the following right triangle: From this we see that sec θ = u +9. Thus u + 9 I = ln + u + C 1 = ln u u + (ln 1 + C 1) = ln (x + ) x + + C Problem x ln(x + 1) SOLUTION. Start by using integration by parts with u = ln(x + 1), dv = x, then du = x x +1 and v = 1 x, so that By long division, I = x ln(x + 1) = 1 x ln(x + 1) x x + 1
5 , so x x + 1 = x x + 1 x x + 1 = (x x + 1 ) = 1 x x + tan 1 x + C Therefore, I = 1 x ln(x + 1) (1 x x + tan 1 x) + C Problem Find the volume of the solid obtained by revolving the graph of y = x 1 x over [, 1] about the y-axis. SOLUTION. Using the method of cylindrical shells, the volume is given by V = π 1 x(x 1 x ) = π 1 x 1 x To evaluate this, let x = sin θ, = cos θdθ so x 1 x = sin θ 1 sin θ cos θdθ = sin θ cos dθ = cos θ cos θ dθ Using the reduction formula for cos θdθ, cos θ cos θ dθ = = 1 cos θ sin θ + 1 [ cos cos θ sin θ θdθ + ] cos θ dθ = 1 cos θ sin θ+ 1 cos θdθ [ 1 θ + 1 ] sin θ cos θ + C = 1 cos θ sin θ θ + 1 sin θ cos θ + C 8 Since sin θ = x, we use the following triangle to see that cos θ = 1 x Thus, cos θ cos θ dθ = 1 (1 x ) / x arcsin x x 1 x + C 5
6 Using this, we compute the volume: V = π ( 1 x(1 x ) / + 18 arcsin x + 18 ) 1 x 1 x = π [( + π ) ] 16 + () = π 8 Problem Find the volume of the solid obtained by revolving the region between the graph of y x = 1 and the line y = about the line y =. SOLUTION. First solve the equation y x = 1 for y: y = ± x + 1 The region in question is bounded in part by the top half of this hyperbola, which is the equation. y = x + 1 The limits of integration are obtained by finding the points of intersection of this equation with y = : = x + 1 x = ± The radius of each disk is given by x + 1; the volume is therefore given by = 8π V = πr = π π 8π ( x + 1) = [ x (x + 1)] x π (x + 1) To evaluate the integral x + 1, let x = tanθ. Then = sec θdθ, x + 1 = sec θ x + 1 = sec θdθ = 1 tanθsecθ + 1 secθdθ = 1 tanθsecθ + 1 ln secθ + tanθ + C = 1 x x ln x x + C 6
7 Now we can compute the volume: [ V = 8πx 8π( 1 x x ln x x ) + ] πx + πx = π[ ln + ] Problem 8.5. Find the constants in the partial fraction decomposition SOLUTION. Clearing denominators gives: x + (x )(x + ) = A x + Bx + C x + x + = A(x + ) + (Bx + C)(x ) 8..9 Setting x = then yields: A = 1 To find B and C, expand the right side, gather like terms, and use the method of undetermined coefficients: x + = (B + 1)x + ( B + C)x + ( C) Equating x coefficients we find B = 1. While equating constants yields C =. Problem Evaluate the integral. (x + x ) (x + )(x + 5)(x ) SOLUTION. The partial fraction decomposition has the form: 8.5. Clearing denominators gives us (x + x ) (x + )(x + 5)(x ) = A x + + B x C x x + x = A(x + 5)(x ) + B(x + )(x ) + C(x + )(x + 5). Setting x = then yields A =, while setting x = 5 yields B = 1 and setting x = yields C =. The result is: (x + x ) (x + )(x + 5)(x ) = x x x Thus (x + x ) (x + )(x + 5)(x ) = 7 x + x + 5 x
8 = ln x + ln x + 5 ln x + C Problem Evaluate x 1/ x 1/. Hint: Use the substitution u = x 1/ SOLUTION. By long division and substitution u = x 1/6, du = 1 6 x 5/6 6x 5/6 du = 6u 5 du = x 1/ x 1/ = 6u 5 u u du = 6 u ( u 1 du = 6 u + u ) du = u 1 ( 1 6 u + 1 ) u + u + ln u 1 +C = u +u +6u+6 ln u 1 +C = x 1/ +x 1/ +6x 1/6 +6 ln x 1/6 1 +C Problem Use the substitution of Exercise 57 to evaluate dθ cosθ+sinθ SOLUTION. Using the substitution θ = tan 1 t, we get dθ cosθ + sinθ = dt (1+t ) 1 t 1+t + t 1+t = The partial fraction decomposition has the form dt 1 t + t = dt t t 1 Clearing denominators gives us t t 1 = A t 1 + B t 1 + = A(t 1 + ) + B(t 1 ) Setting t = 1 + then yields A = 1, while setting t = 1 yields B = 1. Thus, dθ cosθ + sinθ = 1 dt t dt t 1 = 1 ln t ln t 1 + C = 1 ln tan( θ ) 1 + tan( θ ) 1 + C
9 Problem Suppose that Q(x) = (x a 1 )(x a )...(x a n ), where the roots a j are all distinct. Let P Q be a proper rational function so that (a) Show that A j = P(a j) Q (a j ) for j = 1,..., n. P(x) Q(x) = A 1 x a 1 + A x a A n x a n (b) Use this result to find the partial fraction decomposition for P(x) = x 1, Q(x) = x x + x + 6 = (x + 1)(x )(x ). SOLUTION. To differentiate Q(x), first take the logarithm of both sides, and then differentiate: ln(q(x)) = ln[(x a 1 )(x a )...(x a n )] = ln(x a 1 ) + ln(x a ) ln(x a n ) d ln(q(x)) = Q (x) Q(x) = x a 1 x a Multiplying both sides by Q(x) gives us Q 1 1 (x) = Q(x)[ ] x a 1 x a n 1 x a n = (x a )(x a )...(x a n ) + (x a 1 )(x a )...(x a n ) (x a 1 )(x a )...(x a n 1 ) In other words, the ith product in the formula for Q (x) has the (x a i ) factor removed. This means that Q (a j ) = (a j a 1 )...(a j a j 1 )(a j a j+1 )...(a j a n ). Now clear the denominators in the expression for P(x) Q(x) : P(x) = A 1Q(x) x a A nq(x) x a n = A 1 (x a )...(x a n ) + (x a 1 )A (x a )...(x a n ) (x a 1 )(x a )...(x a n 1 )A n Setting x = a j, we get So that P(a j ) = (a j a 1 )(a j a )...(a j a j 1 )A j (a j a j+1 )...(a j a n )) A j = P(a j ) (a j a 1 )...(a j a j 1 )(a j a j+1 )...(a j a n ) = P(a j) Q (a j ) 9
10 (b) Let P(x) = x 1 and Q(x) = (x + 1)(x )(x ), so that Q (x) = x 8x + 1. Then a 1 = 1, a =, and a =, so that ; A 1 = P( 1) Q ( 1) = 1 1 ; A = P() Q () = 7 ; A = P() Q () = 17 Thus P(x) Q(x) = 1 1(x + 1) 7 (x ) + 17 (x ) 1
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