#$&* course MTH 174 10/26/12 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Int 1/2(1-cos 2x) dx 1/2 int (1 - cos 2x) dx 1/2 (x - sin(2x)) + C 1/2 (x - sin(x) cos(x) ) + C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Good student solution: The answer is -1/2 (sinx * cosx) + x/2 + C I arrived at this using integration by parts: u= sinx u' = cosx v'= sinx v = -cosx int(sin^2x)= sinx(-cosx) - int(cosx(-cosx)) int(sin^2x)= -sinx(cosx) +int(cos^2(x)) cos^2(x) = 1-sin^2(x) therefore int(sin^2x)= -sinx(cosx) + int(1-sin^2(x)) int(sin^2x)= -sinx(cosx) + int(1) - int(sin^2(x)) 2int(sin^2x)= -sinx(cosx) + int(1dx) 2int(sin^2x)= -sinx(cosx) + x int(sin^2x)= -1/2 sinx(-cosx) + x/2 INSTRUCTOR COMMENT: This is the appropriate method to use in this section. You could alternatively use trigonometric identities such as sin^2(x) = (1 - cos(2x) ) / 2 and sin(2x) = 2 sin x cos x. Solution by trigonometric identities: sin^2(x) = (1 - cos(2x) ) / 2 so the antiderivative is 1/2 ( x - sin(2x) / 2 ) + c = 1/2 ( x - sin x cos x) + c. note that sin(2x) = 2 sin x cos x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: Section 7.2 Problem 4 problem 7.2.4 (previously 7.2.16 was 7.3.18) antiderivative of (t+2) `sqrt(2+3t) **** what is the requested antiderivative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: U = t+2 Du = 1 Dv = (2+3t)^1/2 Int(2+3t)^1/2 U = 2+3t Du = 3 dt Dt = du/3 V = 2/9(2+3t)^3/2 (t+2)(2/9(2+3t)^3/2)) - int(2/9(2+3t)^3/2) dt Int(2/9(2+3t)^3/2) dt U = 2+3t Du = 3 dt Dt = du/3 2/9 int(u^3/2) du/3 2/27 int(u^3/2) 2/27 (1/(5/2)u^5/2) 4/135(2+3t)^5/2 (t+2)(2/9(2+3t)^3/2) - 4/135(2+3t)^5/2 + C (2+3t)^3/2 (30/135(t+2) - 4/135(3t+2) 2(9t + 26)(3t+2)^3/2 / 135 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you use u=t+2 u'=1 v'=(2+3t)^(1/2) v=2/9 (3t+2)^(3/2) then you get 2/9 (t+2) (3t+2)^(3/2) - integral( 2/9 (3t+2)^(3/2) dt ) or 2/9 (t+2) (3t+2)^(3/2) - 2 / (3 * 5/2 * 9) (3t+2)^(5/2) or 2/9 (t+2) (3t+2)^(3/2) - 4/135 (3t+2)^(5/2). Factoring out (3t + 2)^(3/2) you get (3t+2)^(3/2) [ 2/9 (t+2) - 4/135 (3t+2) ] or (3t+2)^(3/2) [ 30/135 (t+2) - 4/135 (3t+2) ] or (3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135 which simplifies to 2( 9t + 26) ( 3t+2)^(3/2) / 135.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: Section 7.2 Problem 8 **** problem 7.2.8 (previously 7.2.27 was 7.3.12) antiderivative of x^5 cos(x^3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: U = x^3 Du = 3x^2 Dv = x^2 cos(x^3) V = 1/3 sin (x^3) X^3(1/3 sin (x^3)) - int(1/3 sin(x^3) 3x^2 dx 1/3(x^3 sin x^3) - 1/3 int(sin (x^3) 3x^2) dx U = x^3 Du = 3x^2 du Dx = du/3x^2 1/3(x^3 sin (x^3)) - 1/3 int(sin u) 3x^2 du/3x^2 1/3(x^3 sin(x^3)) + 1/3 cos (u) + C 1/3(x sin (x^3) + cos (u)) + C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It usually takes some trial and error to get this one: We could try u = x^5, v ' = cos(x^3), but we can't integrate cos(x^3) to get an expression for v. We could try u = cos(x^3) and v' = x^5. We would get u ' = -3x^2 cos(x^3) and v = x^6 / 6. We would end up having to integrate v u ' = -x^8 / 18 cos(x^3), and that's worse than what we started with. We could try u = x^4 and v ' = x cos(x^3), or u = x^3 and v ' = x^2 cos(x^3), or u = x^2 and v ' = x^3 cos(x^3), etc.. The combination that works is the one for which we can find an antiderivative of v '. That turns out to be the following: Let u = x^3, v' = x^2 cos(x^3). Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have 1/3 * x^3 sin(x^3) - 1/3 * int(3 x^2 sin(x^3) dx). Now let u = x^3 so du/dx = 3x^2. You get 1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ). It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you come across the right one. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: **** What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I first used integration by parts and then applied substitution to the right-hand side terms to arrive at the final answer. I attempted multiple values for u and dv before arriving at the answer. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: TYPICAL STUDENT COMMENT: I tried several things: v'=cos(x^3) v=int of v' u=x^5 u'=5x^4 I tried to figure out the int of cos(x^3), but I keep getting confused: It becomes the int of 1/3cosudu/u^(1/3) I feel like I`m going in circles with some of these. INSTRUCTOR RESPONSE: As noted in the given solution, it often takes some trial and error. With practice you learn what to look for.
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00:53:03 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: **** What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I first used integration by parts and then applied substitution to the right-hand side terms to arrive at the final answer. I attempted multiple values for u and dv before arriving at the answer. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: TYPICAL STUDENT COMMENT: I tried several things: v'=cos(x^3) v=int of v' u=x^5 u'=5x^4 I tried to figure out the int of cos(x^3), but I keep getting confused: It becomes the int of 1/3cosudu/u^(1/3) I feel like I`m going in circles with some of these. INSTRUCTOR RESPONSE: As noted in the given solution, it often takes some trial and error. With practice you learn what to look for.
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00:53:03 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok #*&!