12 Z cos(2x)dx =−1 2x cos(2x

10.15 1) f⁰(x) = sin(2x) f(x) =

Z

sin(2x)dx= Z

sin(2x)·2·¹₂ dx= ¹₂ Z

sin(2x)·2dx

= ¹₂ −cos(2x)

=−¹₂ cos(2x) g(x) = x

g⁰(x) = 1 Z

xsin(2x)dx =−¹₂ cos(2x)·x− Z

−¹₂ cos(2x)·1dx

=−¹₂x cos(2x) + ¹₂ Z

cos(2x)dx

=−1

2x cos(2x) + 1 2

Z

cos(2x)·2· ¹₂dx

=−¹₂x cos(2x) + ¹₂ · ¹₂ Z

cos(2x)·2dx

=−¹₂x cos(2x) + ¹₄ sin(2x) +c 2) f⁰(x) = e^x f(x) = e^x

g(x) = x g⁰(x) = 1 Z

x e^xdx=x e^x− Z

e^xdx=x e^x−e^x = (x−1)e^x+c 3) f⁰(x) = e^3x

f(x) = Z

e^3xdx= Z

e^3x·3· ¹₃dx= ¹₃ Z

e^3x·3dx= ¹₃e^3x g(x) = 3x² g⁰(x) = 6x

Z

3x²e^3xdx= ¹₃e^3x·3x² − Z

1

3e^3x·6x dx=x²e^3x− Z

2x e^3xdx Pour calculer

Z

2x e^3xdx, on fait à nouveau une intégration par parties : f⁰(x) = e^3x f(x) = ¹₃e^3x

g(x) = 2x g⁰(x) = 2 Z

2x e^3xdx= ¹₃e^3x·2x− Z

1

3e^3x·2dx= ²₃x e^3x− ²₃ Z

e^3xdx

= ²₃x e^3x−²₃ · ¹₃e^3x = ²₃x e^3x−²₉e^3x Finalement,

Z

3x²e^3xdx=x²e^3x− ²₃x e^3x− ²₉e^3x

=x²e^3x−²₃ x e^3x+²₉e^3x

= ¹₉(9x²−6x+ 2)e^3x+c

Analyse : primitives Corrigé 10.15

4) f⁰(x) = cos(x) f(x) = sin(x) g(x) = x²+ 1 g⁰(x) = 2x Z

(x²+ 1) cos(x)dx= (x²+ 1) sin(x)− Z

2x sin(x)dx Pour calculer

Z

2x sin(x)dx, on procède à une nouvelle intégration par parties :

f⁰(x) = sin(x) f(x) = −cos(x) g(x) = 2x g⁰(x) = 2

Z

2x sin(x)dx =−2x cos(x)− Z

−2 cos(x)dx=−2x cos(x) + 2 sin(x) Donc

Z

(x²+ 1) cos(x)dx= (x²+ 1) sin(x)− −2x cos(x) + 2 sin(x)

= (x²−1) sin(x) + 2x cos(x) +c 5) f⁰(x) = 1 f(x) =x

g(x) = ln(x) g⁰(x) = 1 x Z

ln(x)dx=x ln(x)− Z

x· 1

xdx=x ln(x)− Z

1dx=x ln(x)−x

=x ln(x)−1 +c 6) f⁰(x) = √

x+ 1 f(x) =

Z √

x+ 1dx= Z

(x+ 1)¹² dx= 1

3 2

(x+ 1)³² = ²₃p

(x+ 1)³

= ²₃(x+ 1)√ x+ 1 g(x) = x g⁰(x) = 1 Z

x√

x+ 1dx= ²₃x(x+ 1)√

x+ 1− Z

2

3 (x+ 1)√

x+ 1dx

= ²₃x(x+ 1)√

x+ 1−²₃ Z

(x+ 1)³² dx

= ²₃x(x+ 1)√

x+ 1−²₃ · 1

5 2

(x+ 1)⁵²

= ²₃x(x+ 1)√

x+ 1−₁₅⁴ (x+ 1)²√ x+ 1

= ₁₅² (x+ 1)√

x+ 1 5x−2 (x+ 1)

= ₁₅² (3x−2) (x+ 1)√

x+ 1 +c

Analyse : primitives Corrigé 10.15

7) f⁰(x) = 1 f(x) =x g(x) = arcsin(x) g⁰(x) = 1

√1−x² Z

arcsin(x)dx=x arcsin(x)−

Z 1

√1−x² ·x dx

=x arcsin(x)−(−¹₂) Z

(1−x²)⁻¹² ·(−2x)dx

=x arcsin(x) + ¹₂ · 1

1 2

(1−x²)¹²

=x arcsin(x) +√

1−x²+c 8) f⁰(x) = cos(x) f(x) = sin(x)

g(x) = e^x g⁰(x) =e^x Z

e^x cos(x)dx=e^x sin(x)− Z

e^x sin(x)dx Pour calculer

Z

e^x sin(x)dx, on recourt derechef à une intégration par parties :

f⁰(x) = sin(x) f(x) = −cos(x) g(x) = e^x g⁰(x) =e^x Z

e^x sin(x)dx=−e^x cos(x)−

Z

−e^x cos(x)dx =−e^x cos(x)+

Z

e^x cos(x)dx Ainsi

Z

e^x cos(x)dx=e^x sin(x)−

−e^x cos(x) + Z

e^x cos(x)dx

=e^x sin(x) + cos(x)

− Z

e^x cos(x)dx Il en résulte 2

Z

e^x cos(x)dx=e^x sin(x) + cos(x) d’où finalement

Z

e^x cos(x)dx= ¹₂e^x sin(x) + cos(x) +c 9) f⁰(x) = x f(x) = ¹₂x²

g(x) = ln(x) g⁰(x) = 1 x Z

xln(x)dx= ¹₂x² ln(x)− Z

1 2x² · 1

xdx= ¹₂x² ln(x)− ¹₂ Z

x dx

= ¹₂x² ln(x)−¹₄ x² = ¹₄x² 2 ln(x)−1 +c

Analyse : primitives Corrigé 10.15

10) f⁰(x) = e^−x f(x) =−e^−x g(x) = x² g⁰(x) = 2x Z

x²e^−xdx =−x²e^−x− Z

−2x e^−xdx=−x²e^−x+ 2 Z

x e^−xdx

Pour calculer Z

x e^−xdx, on procède encore par intégration par parties : f⁰(x) = e^−x f(x) =−e^−x

g(x) = x g⁰(x) = 1 Z

x e^−xdx=−x e^−x− Z

−e^−xdx=−x e^−x−e^−x D’où

Z

x²e^−xdx=−x²e^−x+ 2 −x e^−x−e^−x

=−(x²+ 2x+ 2)e^−x+c

Analyse : primitives Corrigé 10.15