# quiz 2 - 2 18 x 3 = Ax B x 2 9 Cx D x 2 1 = A C x 3 B D x...

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Math 1B: Calculus February 3, 2010 Quiz 2 Lecturer: Prof. Mina Aganagic GSI: Gary Sivek Name: Answers 1. (2 pts) Evaluate ± dx (4 - x 2 ) 3 / 2 . We have the expression a 2 - x 2 in the integrand (even if cubed), with a =2 ,so we make the substitution x = a sin θ =2s in θ , dx =2cos θdθ : ± dx (4 - x 2 ) 3 / 2 = ± 2cos θdθ (4 - 4sin 2 θ ) 3 / 2 = ± 2cos θdθ (4cos 2 θ ) 3 / 2 = ± 2cos θdθ 8cos 3 θ = ± 1 4 sec 2 θdθ = 1 4 tan θ + C. Now tan θ = sin θ cos θ = sin θ ² 1 - sin 2 θ = x/ 2 ² 1 - ( x/ 2) 2 = x 4 - x 2 ,so : ± dx (4 - x 2 ) 3 / 2 = x 4 4 - x 2 + C. 2. (3 pts) Evaluate ± 2 x 3 +3 x 2 +18 x +3 x 4 +10 x 2 +9 dx .T h ed e n om in a t o ri s( x 2 +1)( x 2 +9), so write Ax + B x 2 +1 + Cx + D x 2 +9 = 2 x 3 +3 x 2 +18
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Unformatted text preview: 2 + 18 x + 3 = ( Ax + B )( x 2 + 9) + ( Cx + D )( x 2 + 1) = ( A + C ) x 3 + ( B + D ) x 2 + (9 A + C ) x + (9 B + D ) From A + C = 2 and 9 A + C = 18 we get A = 2 and C = 0; from B + D = 3 and 9 B + D = 3 we get B = 0 and D = 3. So the integral becomes: ± 2 x x 2 + 1 dx + ± 3 x 2 + 9 dx = ± du u + 3 ± dx x 2 + 9 ( u = x 2 + 1 , du = 2 x dx ) = ln | u | + 3 · 1 3 tan-1 x 3 + C = ln ( x 2 + 1) + tan-1 x 3 + C....
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