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33-substitution-day-2

# 33-substitution-day-2 - using u = 2 x 41 This is 1 2 Z 65 1...

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Math 1a More Substitution Fall, 2009 31 Z x 2 dx x 3 + 1 32 Z xe x 2 dx 33 Z dx x (ln x ) 2 + 4 x ln x + 4 x 34 Z e x x dx 35 Z 1 - 1 ( x + 1) 3 dx 36 Z π/ 9 0 sin(3 x ) dx 37 Z 3 0 dx (2 x + 1) 2 38 Z π/ 2 π/ 4 cot( x ) dx

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39 Z π/ 6 0 sin 3 (2 x ) cos(2 x ) dx 40 Z 1 0 1 1 + 4 x 2 dx 41 Z 4 0 x 1 + 4 x 2 dx 42 Z 1 0 x 1 + 16 x 4 dx 43 Z 1 - 1 x ( x 2 + 1) 3 dx 44 Z 4 1 ( x + 3) 3 x dx
Math 1a More Substitution – Answers Fall, 2009 31 2 3 x 3 + 1 + C using u = x 3 + 1 32 1 2 e x 2 + C using u = x 2 33 This becomes Z du u 2 + 4 u + 4 using u = ln x , so the answer is - 1 ln( x ) + 2 + C 34 2 e x + C using u = x 35 This is Z 2 0 u 3 du = 4 36 This is 1 3 Z π/ 3 0 sin( u ) du = 1 6 37 This is 1 2 Z 7 1 u - 2 du = 1 2 1 - 1 7 = 3 n 7 using u = 2 x + 1 38 This is Z 1 1 / 2 1 u du = ln(2) 2 using u = sin( x ). (Note question change!) 39 This is 1 2 Z 3 / 2 0 u 3 du = 1 8 3 2 ! 4 = 9 128 using u = sin(2 x ) 40 This is Z 2 0 1 1 + u 2 du
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Unformatted text preview: using u = 2 x 41 This is 1 2 Z 65 1 1 u du = 1 2 ln(65) using u = 1 + 4 x 2 42 This is 1 8 Z 4 1 1 + u 2 du = 1 8 arctan(4) using u = 4 x 2 43 This is 1 2 Z u 3 du = 0 using u = x 2 + 1. Note that this is an odd function on [-1 , 1], so the integral vanishes. 44 This is 2 Z 2 1 ( u + 3) 3 du = 1 2 ( 5 4-4 4 ) = 369 2 using u = √ x...
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