32-substitution-day-1

# 32-substitution-day-1 - Math 1a Substitution Fall 2009 \$...

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Math 1a Substitution Fall, 2009 & \$ % The Substitution Rule: Suppose u = g ( x ) is a differentiable function with domain an interval I and f ( x ) is continuous on I . Then Z f ( g ( x )) g 0 ( x ) dx = Z f ( u ) du. Thus we can make substitutions and treat du and dx like differentials: du = du dx dx . Some Integration Formulas you might like to have at your fingertips: Z u n du = u n +1 n + 1 + C n 6 = - 1 Z 1 u du = ln | u | + C Z e u du = e u + C Z sin( u ) du = - cos( u ) + C Z cos( u ) du = sin( u ) + C Compute the following indefinite integrals: 1 Z ( x 2 + 2 x + 1) dx 2 Z ( x + 1) 2 dx 3 Z (2 x + 1) 2 dx 4 Z x 2 ( x 3 - 4 ) 6 dx 5 Z x x 2 + 1 dx 6 Z x x 2 + 1 dx 7 Z 1 (2 x + 3) 4 dx 8 Z sin(4 x - 3) dx 9 Z x cos( x 2 + 1) dx 10 Z e 4 x +1 dx 11 Z xe x 2 +1 dx 12 Z e x x dx 13 Z 1 2 x dx 14 Z ln( x ) x dx 15 Z 1 x ln( x ) dx

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Some More Integration Formulas that might help with some of the integrals on this page: Z sec 2 ( u ) du = tan( u ) + C Z sec( u ) tan( u ) du = sec( u ) + C Z 1 1 + u 2 du = tan - 1 ( u ) + C Z 1 1 - u 2 du = sin - 1 ( u ) + C Compute the following indefinite integrals: 16 Z 2 x 1 + x 2 dx 17 Z 2 x 1 + x 4 dx 18 Z x 1 - x 2 dx 19 Z 2 x sec 2 ( x 2 ) dx 20 Z sec( x ) tan( x ) x dx 21 Z x + x + 1 x dx 22 Z x 3 1 + x 4 dx 23 Z x 3 1 + x 2 dx 24 Z 1 x (1 + x ) dx 25 Z sin( x ) cos( x ) dx 26 Z tan(5 x ) dx 27 Z sec 2 ( x ) + sec( x ) tan( x ) tan( x ) + sec( x ) dx 28 Z sec( x ) dx 29 Z tan( x ) ( 1 - sec 2 ( x ) ) dx 30 Z tan 3 ( x ) dx
Math 1a Substitution – Answers Fall, 2009 1 1 3 x 3 + x 2 + x + C 2 1 3 ( x + 1) 3 + C 3 1 6 (2 x + 1) 3 + C 4 1 21 ( x 3 - 4) 7 + C 5 1 3 ( x 2 + 1) 3 / 2 + C 6 x 2 + 1 + C 7 - 1 6 (2 x + 3) - 3 + C = -
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