Answers submitted 4tarctan4t ln16tˆ2124 correct

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Answer(s) submitted: (4tarctan(4t)-ln(16tˆ2+1)/2)/4 (correct) Correct Answers: t*atan(4*t)-0.125*ln(1+(4*t)ˆ2) 6. (1 point) Evaluate the indefinite integral. Z e 5 x sin ( 3 x ) dx Answer = + C Solution: SOLUTION For R e 5 x sin ( 3 x ) dx , we use integration by parts: R udv = uv - R vdu . For our first choice of u and dv , we are free to have u = e 5 x and dv = sin ( 3 x ) dx , or u = sin ( 3 x ) and dv = e 5 x dx . We proceed with the latter: Let u = sin ( 3 x ) dv = e 5 x dx then du = 3cos ( 3 x ) dx v = 1 5 e 5 x And so R udv = uv - R vdu R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 5 R e 5 x cos ( 3 x ) dx We now must use integration by parts again, with our choice of u and dv consistent with our original choice. That is, we let u = cos ( 3 x ) dv = e 5 x dx then du = - 3sin ( 3 x ) dx v = 1 5 e 5 x And so R udv = uv - R vdu R e 5 x cos ( 3 x ) dx = 1 5 e 5 x cos ( 3 x )+ 3 5 R e 5 x sin ( 3 x ) dx We substitute this result R e 5 x cos ( 3 x ) dx = 1 5 e 5 x cos ( 3 x )+ 3 5 R e 5 x sin ( 3 x ) dx into our earlier result: R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 5 R e 5 x cos ( 3 x ) dx to get R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 5 R e 5 x cos ( 3 x ) dx R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 5 1 5 e 5 x cos ( 3 x )+ 3 5 R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 25 e 5 x cos ( 3 x ) - 9 25 R e 5 x sin ( 3 x ) dx + 9 25 R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 25 e 5 x cos ( 3 x ) ( 1 + 9 25 ) R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 25 e 5 x cos ( 3 x ) 34 25 R e 5 x sin ( 3 x ) dx = 1 5 e 5 x sin ( 3 x ) - 3 25 e 5 x cos ( 3 x ) R e 5 x sin ( 3 x ) dx = 5 34 e 5 x sin ( 3 x ) - 3 34 e 5 x cos ( 3 x )+ C R e 5 x sin ( 3 x ) dx = e 5 x 5 34 sin ( 3 x ) - 3 34 cos ( 3 x ) + C Note: if we used u = e 5 x dv = sin ( 3 x ) dx so du = 5 e 5 x dx v = - 1 3 cos ( 3 x ) for our first integration by parts, and 2
u = e 5 x dv = cos ( 3 x ) dx so du = 5 e 5 x dx v = 1 3 sin ( 3 x ) for the second integration by parts, we would have gotten: R e 5 x sin ( 3 x ) dx = - 1 3 e 5 x cos ( 3 x )+ 5 3 R e 5 x cos ( 3 x ) dx R e 5 x sin ( 3 x ) dx = - 1 3 e 5 x cos ( 3 x )+ 5 3 1 3 e 5 x sin ( 3 x ) - 5 3 R e 5 x sin ( 3 x ) dx R e 5 x sin ( 3 x ) dx = - 1 3 e 5 x cos ( 3 x )+ 5 9 e 5 x sin ( 3 x ) - 25 9 R e 5 x sin ( 3 x ) dx R e 5 x sin ( 3 x ) dx + 25 9 R e 5 x sin ( 3 x ) dx = - 1 3 e 5 x cos ( 3 x )+ 5 9 e 5 x sin ( 3 x ) ( 1 + 25 9 ) R e 5 x sin ( 3 x ) dx = - 1 3 e 5 x cos ( 3 x )+ 5 9 e 5 x sin ( 3 x ) 34 9 R e 5 x sin ( 3 x ) dx = - 1 3 e 5 x cos ( 3 x )+ 5 9 e 5 x sin ( 3 x ) R e 5 x sin ( 3 x ) dx = - 3 34 e 5 x cos ( 3 x )+ 5 34 e 5 x sin ( 3 x )+ C R e 5 x sin ( 3 x ) dx = e 5 x - 3 34 cos ( 3 x )+ 5 34 sin ( 3 x ) + C The key is that the choices in the second integration by parts must be consistent with the first set of choices. Answer(s) submitted: (eˆ(5x)(5sin(3x)-3cos(3x)))/34 (correct) Correct Answers: -3/34 * (eˆ(5*x) * cos(3*x)) + 5/34 * (eˆ(5*x) * sin(3*x)) 7. (1 point) Evaluate the integral. Z 5 1 ln x x 7 dx Solution: SOLUTION For R 5 1 ln x x 7 dx , we use integration by parts (definite integral version): R b a udv = uv b a - R b a vdu . Let u = ln ( x ) dv = x - 7 dx then du = 1 x dx v = 1 - 6 x - 6 And so R b a udv = uv b a - R b a vdu R 5 1 ln x x 7 dx = - 1 6 ln x x 6 5 1 + 1 6 R 5 1 ( 1 x ) 1 x 6 dx = - 1 6 ( ln5 5 6 - ln1 1 ) + 1 6 R 5 1 x - 7 dx = - 1 6 ( ln5 5 6 ) - 1 36 x - 6 5 1 = - 6 36 ( ln5 5 6 ) - 1 36 h 1 5 6 - 1 i = 1 36 - 6ln5 + 1 ( 36 )( 5 6 ) Answer(s) submitted: (1/36)-((8ln5+1)/(562500)) (correct) Correct Answers: 0.0277588326622674 8. (1 point) Use integration by parts to evaluate the integral.

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