Solucionario Capitulo 7 Calculo Se Stewart

Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts

1. Let u=ln x , dv=xdx xln xdx =

du=dx/x , v=

1 2 x ln x 2 1 2 1 2 = x ln x x +C 2 4
2

1 2 x . Then by Equation 2, udv=uv vdu , 2 1 2 1 2 1 1 2 1 1 2 x ( dx/x ) = x ln x xdx= x ln x x +C 2 2 2 2 2 2

2. Let u= , dv=sec sec
2

d

du=d , v=tan

. Then +C .

d = tan

tan d = tan du=dx ,v=

ln sec

3. Let u=x , dv=cos 5xdx xcos 5xdx= 1 xsin 5x 5
x

1 sin 5x . Then by Equation 2, 5 1 1 1 sin 5xdx= xsin 5x+ cos 5x+C . 5 5 25
x

4. Let u=x , dv=e dx 5. Let u=r , dv=e dr 6. Let u=t , dv=sin 2t dt dt=
r/2

du=dx , v= e du=dr , v=2e du=dt , v=

. Then xe dx= xe + e dx= xe . Then re dr=2re
r/2 r/2 r/2

x

x

x

x

e +C .
r/2

x

r/2

2e dr=2re

4e +C.

r/2

1 1 1 cos 2t . Then tsin 2t dt= tcos 2t+ cos 2t 2 2 2

1 1 tcos 2t+ sin 2t+C . 2 4
2

7. Let u=x , dv=sin xdx I= x sin xdx=
2

du=2xdx and v= 2

1

cos x . Then

1

x cos x+

2

xcos xdx ( * ). 1 sin x , so 1
2

Next let U =x , dV =cos xdx xcos xdx= 1 xsin x 1

dU =dx , V = sin xdx= 1

xsin x+

cos x+C . Substituting for xcos xdx in (
1

* ), we get 1 22 I= x cos x+ where C= 2 C .
1

1

xsin x+

1
2

cos x+C

1

=

1

x cos x+

2

2
2

xsin x+

2
3

cos x+C ,

1

Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts

8. Let u=x , dv=cos mxdx I= x cos mxdx=
2

2

du=2xdx , v=

1 sin mx . Then m

1 2 2 x sin mx xsin mxdx ( * ). Next let U =x , dV =sin mxdx dU =dx , mm 1 1 1 1 1 V= cos mx , so xsin mxdx= xcos mx+ cos mxdx= xcos mx+ sin mx+C . 1 2 m m m m m Substituting for xsin mxdx in ( * ), we get 1 2 2 1 1 2 1 2 2 I= x sin mx xcos mx+ sin mx+C = x sin mx+ xcos mx sin mx+C , 1 2 2 3 m m m m m m m 2 where C= C . m 1 9. Let u=ln (2x+1) , dv=dx ln (2x+1)dx =xln (2x+1) =xln (2x+1) =
1

du=

2 dx , v=x . Then 2x+1 2x (2x+1) 1 dx=xln (2x+1) dx 2x+1 2x+1 1 1 1dx=xln (2x+1) x+ ln (2x+1)+C 2x+1 2

1 (2x+1)ln (2x+1) x+C 2 du= dx 1 x xdx 1 x
2 2

10. Let u=sin x , dv=dx
2

, v=x . Then sin xdx=xsin x
1/2

1

1

x 1 x
2

dx . Setting

t=1 x , we get dt= 2xdx , so
1 1

=

t

1 1 1/2 1/2 2 dt = 2t +C=t +C= 1 x +C . 2 2

(

)

Hence, sin xdx=xsin x+ 1 x +C . 11. Let u=arctan 4t , dv=dt arctan 4t dt =t arctan 4t =t arctan 4t
52

du= 4t 1+16t
2

4 1+(4t)
2

dt=

4
2

dt , v=t . Then 32t 1+16t
2

1+16t 1 dt=t arctan 4t 8

dt

1 2 ln (1+16t )+C 8

12. Let u=ln p , dv= p dp
2

Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts

du=

1 1 6 1 6 1 5 1 6 1 6 5 dp , v= p . Then p ln pdp= p ln p p dp= p ln p p +C . p 6 6 6 6 36
2

13. First let u= ( ln x) , dv=dx I= (ln x) dx=x(ln x) 2 xln x to get ln xdx=xln x
2 2 2

du=2ln x

1 dx , v=x . Then by Equation 2, x dU =1/xdx , V =x

1 2 dx=x(ln x) 2 ln xdx . Next let U =ln x , dV =dx x x ( 1/x ) dx=xln x dx=xln x x+C . Thus,
1 1

I=x(ln x) 2 xln x x+C 14. Let u=t , dv=e dt more with dv=e dt . I = t 3et
t 3 t

(

) =x(ln x)
2

2

2xln x+2x+C , where C= 2C .
1 t 3 t 3 t

du=3tdt , v=e . Then I= t e dt=t e

3t e dt . Integrate by parts twice

2 t

( 3t2et

6te dt =t e 3t e +6te

t

)

3 t

2 t

t

6e dt

t

= t 3et 3t 2et+6tet 6et+C= t 3 3t 2+6t 6 et+C More generally, if p(t) is a polynomial of degree n in t , then repeated integration by parts shows that t / / / / / / n ( n) t p ( t ) e dt= p(t) p (t)+ p (t) p (t)+ + ( 1 ) p (t) e +C . 15. Firstlet u=sin 3 , dv=e d I= e sin 3 d =
2 2

(

)

du=3cos 3 d , v=

1 2 e . Then 2

1 2 3 2 2 e sin 3 e cos 3 d . Next let U =cos 3 , dV =e d 2 2 1 2 dU = 3sin 3 d , V = e to get 2 1 2 3 2 2 e cos 3 d = e cos 3 + e sin 3 d . Substituting in the previous formula gives 2 2 1 2 3 2 9 2 1 2 3 2 9 I= e sin 3 e cos 3 e sin 3 d = e sin 3 e cos 3 I 2 4 4 2 4 4 13 1 2 3 2 1 2 4 I= e sin 3 e cos 3...