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57320-0136147054_07

# 57320-0136147054_07 - Section 7.2 7.2.1 Let u = 2 3x Then...

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Section 7.2 7.2.1: Let u =2 3 x .T h e n du = 3 dx ,ands o Z (2 3 x ) 4 dx = 1 3 Z (2 3 x ) 4 ( 3) dx = 1 3 Z u 4 du = 1 3 · 1 5 u 5 + C = 1 15 (2 3 x ) 5 + C. 7.2.2: Let u =1+2 x .Th en x = 1 2 ( u 1) and dx = 1 2 du ,sotha t Z 1 (1 + 2 x ) 2 dx = 1 2 Z u 2 du = 1 2 u + C = 1 2(1 + 2 x ) + C. 7.2.3: Let u x 3 4. Then du =6 x 2 dx ,s oth a t Z x 2 (2 x 3 4) 1 / 2 dx = 1 6 Z (2 x 3 4) 1 / 2 · 6 x 2 dx = 1 6 Z u 1 / 2 du = 1 6 · 2 3 u 3 / 2 + C = 1 9 (2 x 3 4) 3 / 2 + C. 7.2.4: Let u =5+2 t 2 h e n du =4 tdt ,andso Z 5 t 5+2 t 2 dt = 5 4 Z 4 t t 2 dt = 5 4 Z 1 u du = 5 4 ln | u | + C = 5 4 ln(5 + 2 t 2 )+ C. 7.2.5: Let u x 2 +3. Then du xdx o Z 2 x (2 x 2 +3) 1 / 3 dx = 1 2 Z (2 x 2 1 / 3 · 4 = 1 2 Z u 1 / 3 du = 1 2 · 3 2 u 2 / 3 + C = 3 4 (2 x 2 2 / 3 + C. 7.2.6: Let u = x 2 h e n du , and therefore Z x sec 2 x 2 dx = 1 2 Z ( sec x 2 ) 2 · 2 = 1 2 Z (sec u ) 2 du = 1 2 tan u + C = 1 2 tan( x 2 C = 1 2 tan x 2 + C. 7.2.7: Let u = y 1 / 2 a t du = 1 2 y 1 / 2 dy h e n Z y 1 / 2 cot y 1 / 2 ± csc y 1 / 2 ± dy Z cot u csc udu = 2csc u + C = y + C. 7.2.8: Let u = π (2 x +1). Then du πdx , and hence Z sin π (2 x +1) dx = 1 2 π Z sin = 1 2 π cos u + C = 1 2 π cos π (2 x +1)+ C. 7.2.9: Let u =1+sin θ h e n du =cos θdθ , and therefore Z (1 + sin θ ) 5 cos = Z u 5 du = 1 6 u 6 + C = 1 6 (1 + sin θ ) 6 + C. 643

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7.2.10: Let u =4+cos2 x .T h e n du = 2sin2 xdx , and thus Z sin2 x 4+cos2 x dx = 1 2 Z x x dx = 1 2 Z 1 u du = 1 2 ln | u | + C = 1 2 ln(4 + cos2 x )+ C. 7.2.11: Let u = cot x h e n du =csc 2 .S o Z e cot x csc 2 = Z e u du = e u + C = e cot x + C =exp( cot x C. 7.2.12: Let u =( x +4) 1 / 2 h e n du = 1 2 ( x 1 / 2 dx h u s Z exp ( ( x 1 / 2 ) ( x 1 / 2 dx =2 Z e u du e u + C =2exp ( x 1 / 2 ± + C. 7.2.13: Let u =ln t h e n du = 1 t dt ,so Z (ln t ) 10 t dt = Z u 10 du = 1 11 u 11 + C = 1 11 (ln t ) 11 + C. 7.2.14: Let u =1 9 t 2 h e n du = 18 tdt . Hence Z t 1 9 t 2 dt = 1 18 Z (1 9 t 2 ) 1 / 2 · ( 18 t ) dt = 1 18 Z u 1 / 2 du = 1 18 · 2 u 1 / 2 + C = 1 9 p 1 9 t 2 + C. 7.2.15: Let u =3 t ,sotha t du dt h e n Z 1 1 9 t 2 dt = 1 3 Z 1 1 9 t 2 · 3 dt = 1 3 Z 1 1 u 2 du = 1 3 arcsin u + C = 1 3 arcsin(3 t C. 7.2.16: Let u =1+ e 2 x h e n du e 2 x dx and thus Z e 2 x 1+ e 2 x dx = 1 2 Z 2 e 2 x e 2 x dx = 1 2 Z 1 u du = 1 2 ln | u | + C = 1 2 ln ( e 2 x ) + C. 7.2.17: Let u = e 2 x h e n du e 2 x dx . Therefore Z e 2 x e 4 x dx = 1 2 Z 2 e 2 x 1+( e 2 x ) 2 dx = 1 2 Z 1 u 2 du = 1 2 arctan u + C = 1 2 arctan ( e 2 x ) + C. 7.2.18: Let u =arctan x h e n du = 1 x 2 dx ,s oth a t Z exp(arctan x ) x 2 dx = Z e u du = e u + C = exp (arctan x C. 644
7.2.19: Let u = x 2 ,sotha t du =2 xdx ,ands o Z 3 x 1 x 4 dx = 3 2 Z 2 x 1 x 4 dx = 3 2 Z 1 1 u 2 du = 3 2 arcsin u + C = 3 2 arcsin ( x 2 ) + C. 7.2.20: Let u =s in2 x .T h e n du =2cos2 , and thus Z sin 3 2 x cos2 = 1 2 Z u 3 du = 1 8 u 4 + C = 1 8 sin 4 2 x + C. The Mathematica 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 and a similar command in Derive .

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57320-0136147054_07 - Section 7.2 7.2.1 Let u = 2 3x Then...

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