Chap5_Sec5

Chap5_Sec5 - u = 1 x 2 du = 2 x dx 2 2 3 2 2 3 2 3 2 2 3 2...

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Unformatted text preview: u = 1 + x 2 du = 2 x dx 2 2 3 / 2 2 3 2 3 / 2 2 3 2 1 1 2 ( 1) x x dx x x dx udu u C x C + = + = = + = + + ∫ ∫ ∫ Example (5.5) We can check that we have the correct answer by using the Chain Rule to differentiate the answer 2 3 2 2 1 2 3 2 2 3 3 2 2 ( 1) ( 1) 2 2 1 d x C x x dx x x + + = ⋅ + ⋅ = + SUBSTITUTION RULE Find ∫ x 3 cos( x 4 + 2) dx s We make the substitution u = x 4 + 2. s This is because its differential is du = 4 x 3 dx , which, apart from the constant factor 4, occurs in the integral. Thus, using x 3 dx = du /4 and the Substitution Rule, we have: s Notice that, at the final stage, we had to return to the original variable x . Example 1 (5.5) 3 4 1 1 4 4 1 4 4 1 4 cos( 2) cos cos sin sin( 2) x x dx u du u du u C x C + = ⋅ = ⋅ = + = + + ∫ ∫ ∫ SUBSTITUTION RULE s Let u = 2 x + 1. s Then, du = 2 dx. s So, dx = du /2. Another possible substitution is Let Then So s Alternatively, observe that u 2 = 2 x + 1....
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This note was uploaded on 05/23/2011 for the course MAC 2311 taught by Professor Noohi during the Fall '08 term at FSU.

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Chap5_Sec5 - u = 1 x 2 du = 2 x dx 2 2 3 2 2 3 2 3 2 2 3 2...

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