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Calculus with Analytic Geometry by edwards & Penney soln ch8

Calculus with Analytic Geometry

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Section 8.2 C08S02.001: Let u = 2 3 x . Then du = 3 dx , and so (2 3 x ) 4 dx = 1 3 (2 3 x ) 4 ( 3) dx = 1 3 u 4 du = 1 3 · 1 5 u 5 + C = 1 15 (2 3 x ) 5 + C. C08S02.002: Let u = 1 + 2 x . Then x = 1 2 ( u 1) and dx = 1 2 du , so that 1 (1 + 2 x ) 2 dx = 1 2 u 2 du = 1 2 u + C = 1 2(1 + 2 x ) + C. C08S02.003: Let u = 2 x 3 4. Then du = 6 x 2 dx , so that x 2 (2 x 3 4) 1 / 2 dx = 1 6 (2 x 3 4) 1 / 2 · 6 x 2 dx = 1 6 u 1 / 2 du = 1 6 · 2 3 u 3 / 2 + C = 1 9 (2 x 3 4) 3 / 2 + C. C08S02.004: Let u = 5 + 2 t 2 . Then du = 4 t dt , and so 5 t 5 + 2 t 2 dt = 5 4 4 t 5 + 2 t 2 dt = 5 4 1 u du = 5 4 ln | u | + C = 5 4 ln(5 + 2 t 2 ) + C. C08S02.005: Let u = 2 x 2 + 3. Then du = 4 x dx , and so 2 x (2 x 2 + 3) 1 / 3 dx = 1 2 (2 x 2 + 3) 1 / 3 · 4 x dx = 1 2 u 1 / 3 du = 1 2 · 3 2 u 2 / 3 + C = 3 4 (2 x 2 + 3) 2 / 3 + C. C08S02.006: Let u = x 2 . Then du = 2 x dx , and therefore x sec 2 x 2 dx = 1 2 ( sec x 2 ) 2 · 2 x dx = 1 2 (sec u ) 2 du = 1 2 tan u + C = 1 2 tan( x 2 ) + C = 1 2 tan x 2 + C. C08S02.007: Let u = y 1 / 2 , so that du = 1 2 y 1 / 2 dy . Then y 1 / 2 cot y 1 / 2 csc y 1 / 2 dy = 2 cot u csc u du = 2csc u + C = 2csc y + C. C08S02.008: Let u = π (2 x + 1). Then du = 2 π dx , and hence sin π (2 x + 1) dx = 1 2 π sin u du = 1 2 π cos u + C = 1 2 π cos π (2 x + 1) + C. C08S02.009: Let u = 1 + sin θ . Then du = cos θ dθ , and therefore (1 + sin θ ) 5 cos θ dθ = u 5 du = 1 6 u 6 + C = 1 6 (1 + sin θ ) 6 + C. C08S02.010: Let u = 4 + cos2 x . Then du = 2sin2 x dx , and thus sin2 x 4 + cos2 x dx = 1 2 2sin2 x 4 + cos2 x dx = 1 2 1 u du = 1 2 ln | u | + C = 1 2 ln(4 + cos2 x ) + C. 1

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C08S02.011: Let u = cot x . Then du = csc 2 x dx . So e cot x csc 2 x dx = e u du = e u + C = e cot x + C = exp( cot x ) + C. C08S02.012: Let u = ( x + 4) 1 / 2 . Then du = 1 2 ( x + 4) 1 / 2 dx . Thus exp ( ( x + 4) 1 / 2 ) ( x + 4) 1 / 2 dx = 2 e u du = 2 e u + C = 2exp ( x + 4) 1 / 2 + C. C08S02.013: Let u = ln t . Then du = 1 t dt , so (ln t ) 10 t dt = u 10 du = 1 11 u 11 + C = 1 11 (ln t ) 11 + C. C08S02.014: Let u = 1 9 t 2 . Then du = 18 t dt . Hence t 1 9 t 2 dt = 1 18 (1 9 t 2 ) 1 / 2 · ( 18 t ) dt = 1 18 u 1 / 2 du = 1 18 · 2 u 1 / 2 + C = 1 9 1 9 t 2 + C. C08S02.015: Let u = 3 t , so that du = 3 dt . Then 1 1 9 t 2 dt = 1 3 1 1 9 t 2 · 3 dt = 1 3 1 1 u 2 du = 1 3 arcsin u + C = 1 3 arcsin(3 t ) + C. C08S02.016: Let u = 1 + e 2 x . Then du = 2 e 2 x dx and thus e 2 x 1 + e 2 x dx = 1 2 2 e 2 x 1 + e 2 x dx = 1 2 1 u du = 1 2 ln | u | + C = 1 2 ln ( 1 + e 2 x ) + C. C08S02.017: Let u = e 2 x . Then du = 2 e 2 x dx . Therefore e 2 x 1 + e 4 x dx = 1 2 2 e 2 x 1 + ( e 2 x ) 2 dx = 1 2 1 1 + u 2 du = 1 2 arctan u + C = 1 2 arctan ( e 2 x ) + C. C08S02.018: Let u = arctan x . Then du = 1 1 + x 2 dx , so that exp(arctan x ) 1 + x 2 dx = e u du = e u + C = exp(arctan x ) + C. C08S02.019: Let u = x 2 , so that du = 2 x dx , and so 3 x 1 x 4 dx = 3 2 2 x 1 x 4 dx = 3 2 1 1 u 2 du = 3 2 arcsin u + C = 3 2 arcsin ( x 2 ) + C. C08S02.020: Let u = sin2 x . Then du = 2cos2 x dx , and thus sin 3 2 x cos2 x dx = 1 2 u 3 du = 1 8 u 4 + C = 1 8 sin 4 2 x + C. 2
The Mathematica 3 . 0 command Integrate[ ( (Sin[2 x] 3 ) Cos[2 x], x ] + C produces exactly the same response, as does the analogous command int( sin(2 x) 3 cos(2 x), x) + C; in Maple V Ver. 5 . 1 and a similar command in Derive 2 . 56.

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