note11

# note11 - Math021 week 11 Example 11.1 Evaluate tan xdx...

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Math021, week 11 Example 11.1 Evaluate Z tan xdx. solution: Let u = cos x so that du = - sin xdx . Then, R tan xdx = R sin x cos x dx = - R du u = - ln u + C = - ln cos x + C. Example 11.2 Evaluate Z sec xdx. solution: Let u = tan x + sec x so that du = (sec 2 x + tan x sec x ) dx . Then, R sec xdx = R sec x (tan x +sec x ) tan x +sec x dx = R (sec 2 x +tan x sec x ) dx ) tan x +sec x = R du u = ln u + C = ln(tan x + sec x ) + C. Example 11.3 Evaluate Z sec 4 xdx. solution: Let u = tan x so that du = sec 2 xdx . Then, R sec 4 xdx = R (1 + tan 2 x ) sec 2 xdx = R (1 + u 2 ) du = u + 1 3 u 3 + C = tan x + 1 3 tan 3 x + C. Integration by Parts 1

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Theorem 11.4 (Integration by part) If u , v are functions, Z u ( x ) v 0 ( x ) dx = u ( x ) v ( x ) - Z u 0 ( x ) v ( x ) dx. proof: For every function u , v , ( uv ) 0 = u 0 v + uv 0 . Thus, u ( x ) v ( x ) = Z u 0 ( x ) v ( x ) dx + Z u ( x ) v 0 ( x ) dx or Z u ( x ) v 0 ( x ) dx = u ( x ) v ( x ) - Z u 0 ( x ) v ( x ) dx. Example 11.5 Evaluate Z xe x dx. solution: Let u ( x ) = x , v 0 ( x ) = e x (so that we may take v ( x ) = e x ). Applying integration by part to the given integral yields R xe x dx = xe x - R e x dx = xe x - e x + C. Example 11.6 Evaluate Z ln xdx. solution: Let u ( x ) = ln x , v 0 ( x ) = 1 (so that we may take v ( x ) = x ). Applying integration by part to the given integral yields R ln xdx = x ln x - R 1 x xdx = x ln x - x + C.
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• Fall '10
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• Math, dx, sinn xdx

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