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Prima metodă de schimbare de variabilă - exerciţii rezolvate (2)

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Prima metodă de schimbare de variabilă - exerciţii rezolvate (2)

Scris de Cristina Vuşcan

Marţi, 04 Decembrie 2012 10:28

Să se calculeze primitivele următoarelor funcţii:

1. $h(x)= sqrt {{e^{x} -1}/{e^{x} +1}}, x in (0; ~ +infty)$ ;

Rezolvare: Avem
$int {~}{~}{ sqrt {{e^{x} -1}/{e^{x} +1}}}{dx}= int {~}{~}{ sqrt {{{(e^{x} -1)}^2}/{e^{2x} -1}}}{dx}=$
$= int {~}{~}{{e^{x} -1}/{ sqrt {e^{2x} -1}}}{dx}= int {~}{~}{{e^{x}}/{ sqrt {e^{2x} -1}}}{dx} - int {~}{~}{1/{ sqrt {e^{2x} -1}}}{dx}$ .
Fie $I_1= int {~}{~}{{e^{x}}/{ sqrt {e^{2x} -1}}}{dx}$ şi $I_2= int {~}{~}{1/{ sqrt {e^{2x} -1}}}{dx}$ .
În continuare
(1) $I_1= int {~}{~}{{e^{x}}/{ sqrt {e^{2x} -1}}}{dx}= int {~}{~}{{(e^{x}) prime }/{ sqrt {{(e^{x})}^2 -1}}}{dx}=$
$= ln {delim{|}{e^{x}+ sqrt {{(e^{x})}^2 -1}}{|}}+crond = ln {(e^{x} + sqrt {e^{2x} -1})}+crond$ ;
(2) $I_2= int {~}{~}{1/{ sqrt {e^{2x} -1}}}{dx}= int {~}{~}{1/{e^{x} sqrt {1- 1/{e^{2x}}}}}{dx}=$
$= int {~}{~}{{e^{ -x}}/{ sqrt {1- e^{ -2x}}}}{dx}= int {~}{~}{{e^{ -x}}/{ sqrt {1- {(e^{ -x})}^2}}}{dx}=$
$= - int {~}{~}{{(e^{ -x}) prime }/{ sqrt {1- {(e^{ -x})}^2}}}{dx}= - arcsin {e^{ -x}}+crond$ .
Din (1) şi (2) rezultă acum:
$int {~}{~}{ sqrt {{e^{x} -1}/{e^{x} +1}}}{dx}= I_1 - I_2= ln {(e^{x} + sqrt {e^{2x} -1})} + arcsin {e^{ -x}}+crond$ .

2. $h(x)= { cos { sqrt {x}}}^2, x in (0; ~ +infty)$ ;

Rezolvare: $int {~}{~}{{ cos { sqrt {x}}}^2}{dx}= int {~}{~}{{1+ cos {2 sqrt {x}}}/2}{dx}= x/2 + 1/2 int {~}{~}{ cos {2 sqrt {x}}}{dx}=$
= x/2 + 1/2 int {~}{~}{{ sqrt {x}}/{ sqrt {x}} cdot cos {2 sqrt {x}}}{dx}= x/2 + 1/2 int {~}{~}{ sqrt {x} cdot ( sin {2 sqrt {x}}) prime }{dx}= x/2 + { sqrt {x}}/2 sin {2 sqrt {x}} - 1/2 int {~}{~}{ ( sqrt {x}) prime cdot sin {2 sqrt {x}}}{dx}=
.

3. $h(x)= { sin {x}}/{ cos {x} sqrt {1+{ sin {x}}^2}}, x in ( - {pi}/2; {pi}/2)$ ;

Rezolvare: $int {~}{~}{{ sin {x}}/{ cos {x} sqrt {1+ { sin {x}}^2}}}{dx}= int {~}{~}{{ sin {x}}/{ cos {x} sqrt {1+1- { cos {x}}^2}}}{dx}=$
$= int {~}{~}{{ sin {x}}/{ cos {x} sqrt {2- { cos {x}}^2}}}{dx}= - int {~}{~}{{ ( cos {x}) prime }/{ cos {x} sqrt {2- { cos {x}}^2}}}{dx}$ .
Aplicăm prima schimbare de variabilă pentru funcţiile varphi : ( - {pi}/2; {pi}/2) right (0;1], ~ f: (0;1] right bbR , unde $varphi (x)= cos {x}, ~ f(t)=1/{t sqrt {2- t^2}}$ .
Pentru t in (0;1] avem:
$int {~}{~}{1/{t sqrt {2- t^2}}}{dt}= int {~}{~}{1/{t^2 sqrt { 2/{t^2} -1}}}{dt}=$
$= int {~}{~}{1/{ t^2 sqrt {{({ sqrt {2}}/t)}^2 - 1}}}{dt}= - 1/{ sqrt {2}} int {~}{~}{{({ sqrt {2}}/t) prime }/{ sqrt {{({ sqrt {2}}/t)}^2 - 1}}}{dt}=$
$= - 1/{ sqrt {2}} ln {delim{|}{{ sqrt {2}}/t + sqrt {{( { sqrt {2}}/t)}^2 -1}}{|}}+crond=$
$= - 1/{ sqrt {2}} ln {({ sqrt {2} + sqrt {2 - t^2}}/t)}+crond$ .
Aşadar, conform primei metode de schimbare de variabilă
$int {~}{~}{{ sin {x}}/{ cos {x} sqrt {1+ { sin {x}}^2}}}{dx}= 1/{ sqrt {2}} ln {({ sqrt {2} + sqrt {2 - { cos {x}}^2}/{ cos {x}})}+crond$ .

4. $h(x)= 1/{(x^2 +1) sqrt {x^2 +2}}, x in (0; ~ +infty)$ ;

Rezolvare: $int {~}{~}{1/{(x^2 +1) sqrt {x^2 +2}}}{dx}= int {~}{~}{x/{(x^2 +1) x sqrt {x^2 +2}}}{dx}=$
$= int {~}{~}{x/{(x^2 +1) sqrt {x^4 +2x^2}}}{dx}= int {~}{~}{{x(x^2 +1)}/{{(x^2 +1)}^2 sqrt {x^4 + 2x^2}}}{dx}=$
$= int {~}{~}{{x^3 +x}/{(x^4 +2x^2 +1) sqrt {x^4 +2x^2}}}{dx}= 1/4 int {~}{~}{{(x^4 +2x^2) prime }/{(x^4 +2x^2 +1) sqrt {x^4 +2x^2}}}{dx}=$
$= 1/2 int {~}{~}{{( sqrt {x^4 +2x^2}) prime }/{{( sqrt {x^4 +2x^2})}^2 +1}}{dx}=$
$= 1/2 arctan { sqrt {x^4 +2x^2}}+crond$ .

5. $h(x)= {x^2 arctan {x}}/{1+x^2}, ~ x in bbR$ ;

Rezolvare: $int {~}{~}{{x^2 arctan {x}}/{1+x^2}}{dx}= int {~}{~}{{(1+x^2 -1) arctan {x}}/{1+x^2}}{dx}=$
$= int {~}{~}{arctan {x}}{dx} - int {~}{~}{{arctan {x}}/{1+x^2}}{dx}= int {~}{~}{(x) prime cdot arctan {x}}{dx} - int {~}{~}{ arctan {x} cdot ( arctan {x}) prime }{dx}=$
$= x arctan {x} - int {~}{~}{x cdot 1/{1+x^2}}{dx} - {{arctan {x}}^2}/2= x arctan {x} - {{arctan {x}}^2}/2 - 1/2 int {~}{~}{{(1+x^2) prime }/{1+x^2}}{dx}=$
$= x arctan {x} - {{arctan {x}}^2}/2 - 1/2 ln {(1+x^2)} + crond$ .

6. $h(x)= 1/{x sqrt {1+x+x^2}}, ~ x in (0; ~ +infty)$ ;

Rezolvare: $int {~}{~}{1/{x sqrt (1+x+x^2}}}{dx}= int {~}{~}{1/{x^2 sqrt {1/{x^2} +1/x +1}}}{dx}=$
$= int {~}{~}{1/{x^2 sqrt {{(1/x)}^2 +1/x +1}}}{dx}= - int {~}{~}{{(1/x + 1/2) prime }/{ sqrt {{(1/x + 1/2)}^2 + 3/4}}}{dx}=$
$= - int {~}{~}{{(1/x + 1/2) prime }/{ sqrt {{(1/x + 1/2)}^2 + {({ sqrt {3}}/2)}^2}}}{dx}= - ln {[(1/x + 1/2) + sqrt {{(1/x + 1/2)}^2 + {({ sqrt {3}}/2)}^2}]}+crond= - ln {({x+2}/{2x} + sqrt {1/{x^2} + 1/x +1})}+crond=$
$= - ln {({x+2+ 2 sqrt {1+x+x^2}}/{2x})}+crond$ .

7. $h(x)= sqrt {x^2 + a^2}, ~ x in bbR, ~ a != 0$ ;

Rezolvare: $int {~}{~}{ sqrt {x^2 + a^2}}{dx}= int {~}{~}{ {x^2 + a^2}/{ sqrt {x^2 + a^2}}}{dx}= int {~}{~}{({x^2}/{ sqrt {x^2 + a^2}} + {a^2}/{ sqrt {x^2 + a^2}})}{dx}=$
$= a^2 ln {(x+ sqrt {x^2 + a^2})} + int {~}{~}{ x cdot ( sqrt {x^2 + a^2}) prime }{dx}= a^2 ln {(x+ sqrt {x^2 + a^2})} + x sqrt {x^2 + a^2} - int {~}{~}{ sqrt {x^2 + a^2}}{dx}$ .
Trecând în membrul stâng cu semn schimbat termenul $int {~}{~}{ sqrt {x^2 + a^2}}{dx}$ , obţinem
$2 int {~}{~}{ sqrt {x^2 + a^2}}{dx}= x sqrt {x^2 + a^2} + a^2 ln {(x+ sqrt {x^2 + a^2})}+crond doubleleftright$
$int {~}{~}{ sqrt {x^2 + a^2}}{dx}= 1/2 [ x sqrt {x^2 + a^2} + a^2 ln {(x+ sqrt {x^2 + a^2})}]+crond$ .

8. $h(x)= sqrt {x^2 - a^2}, ~ x in ( -infty; -a)$ sau x in (a; ~ +infty), ~ a>0 ;

Rezolvare: $int {~}{~}{ sqrt {x^2 - a^2}}{dx}= int {~}{~}{ {x^2 - a^2}/{ sqrt {x^2 - a^2}}}{dx}= int {~}{~}{({x^2}/{ sqrt {x^2 - a^2}} - {a^2}/{ sqrt {x^2 - a^2}})}{dx}=$
$= - a^2 ln {delim{|}{x+ sqrt {x^2 - a^2}}{|}} + int {~}{~}{ x cdot ( sqrt {x^2 - a^2}) prime }{dx}= -a^2 ln {delim{|}{x+ sqrt {x^2 - a^2}}{|}} + x sqrt {x^2 - a^2} - int {~}{~}{ sqrt {x^2 - a^2}}{dx}$ .
Trecând în membrul stâng cu semn schimbat termenul $int {~}{~}{ sqrt {x^2 - a^2}}{dx}$ , obţinem
$2 int {~}{~}{ sqrt {x^2 - a^2}}{dx}= x sqrt {x^2 - a^2} - a^2 ln {delim{|}{x+ sqrt {x^2 - a^2}}{|}}+crond doubleleftright$
$int {~}{~}{ sqrt {x^2 - a^2}}{dx}= 1/2 ( x sqrt {x^2 - a^2} - a^2 ln {delim{|}{x+ sqrt {x^2 - a^2}}{|}})+crond$ .

9. h(x)= 1/{1+ sqrt {x} + sqrt {x+1}}, ~ x in (0; ~ +infty) ;

Rezolvare: $int {~}{~}{1/{1+ sqrt {x} + sqrt {x+1}}}{dx}= int {~}{~}{{(1+ sqrt {x}) - sqrt {x+1}}/{{(1+ sqrt {x})}^2 - {( sqrt {x+1})}^2}}{dx}=$
= int {~}{~}{{1+ sqrt {x} - sqrt {x+1}}/{1+ 2 sqrt {x} +x-x-1}}{dx}= 1/2 int {~}{~}{{1+ sqrt {x} - sqrt {x+1}}/{ sqrt {x}}}{dx}=
= 1/2 int {~}{~}{1/{ sqrt {x}}}{dx} + 1/2 int {~}{~}{1}{dx} - 1/2 int {~}{~}{{ sqrt {x+1}}/{ sqrt {x}}}{dx}=
= sqrt {x} + x/2 - 1/2 int {~}{~}{{ sqrt {x+1}}/{ sqrt {x}}}{dx} (*)
Calculăm
I= int {~}{~}{{ sqrt {x+1}}/{ sqrt {x}}}{dx}= 2 int {~}{~}{ ( sqrt {x}) prime cdot sqrt {x+1}}{dx}=
= 2 sqrt {x} sqrt {x+1} -2 int {~}{~}{ sqrt {x} cdot ( sqrt {x+1}) prime }{dx}= 2 sqrt {x} sqrt {x+1} - int {~}{~}{{ sqrt {x}}/{ sqrt {x+1}}}{dx}=

$= 2 sqrt {x} sqrt {x+1} +2 int {~}{~}{ x^2 ( sqrt {{x+1}/x}) prime }{dx}= 2 sqrt {x} sqrt {x+1} +2x^2 sqrt {{x+1}/x} -2 int {~}{~}{ 2x sqrt {{x+1}/x}}{dx}=$
$= 2 sqrt {x} sqrt {x+1} +2x^2 sqrt {{x+1}/x} -4 int {~}{~}{ sqrt {x^2 +x}}{dx}= 2 sqrt {x} sqrt {x+1} +2x^2 sqrt {{x+1}/x} -4 int {~}{~}{ sqrt {{(x+1/2)}^2 - 1/4}}{dx}$ .
Aplicând rezultatul de la 8, obţinem
$I= 2 sqrt {x} sqrt {x+1} +2x^2 sqrt {{x+1}/x} -2 [(x+1/2) sqrt {x^2 +x} - 1/4 ln {(x+1/2 + sqrt {x^2 +x})}]+crond$ .
Înlocuind în (*) valoarea găsită pentru I, obţinem:
$int {~}{~}{1/{1+ sqrt {x} + sqrt {x+1}}}{dx}= sqrt {x} + x/2 - sqrt {x} sqrt {x+1} - x^2 sqrt {{x+1}/x} + (x+1/2) sqrt {x^2 +x} - 1/4 ln {(x+1/2 + sqrt {x^2 + x})}+crond$ .

10. $h(x)= 1/{(x^2 +2) sqrt {x^2 +1}}, x in (0; ~ +infty)$ ;

Rezolvare: $int {~}{~}{1/{(x^2 +2) sqrt {x^2 +1}}}{dx}= int {~}{~}{1/{[2(x^2 +1) -x^2] sqrt {x^2 +1}}}{dx}=$
$= int {~}{~}{1/{ x^2 [{2(x^2 +1)}/{x^2} -1] sqrt {x^2 +1}}}{dx}= int {~}{~}{1/{{[{ sqrt {2 (x^2 +1)}}/x]}^2 -1} cdot 1/{x^2 sqrt {x^2 +1}}}{dx}=$
$= - 1/{ sqrt {2}} int {~}{~}{1/{{[{ sqrt {2 (x^2 +1)}}/x]}^2 -1} cdot [{ sqrt {2 (x^2 +1)}}/x] prime }{dx}= - 1/{ sqrt {2}} cdot 1/2 ln {delim{|}{{{ sqrt {2 (x^2 +1)}}/x -1}/{{ sqrt {2 (x^2 +1)}}/x +1}}{|}}+crond=$
$= - 1/{2 sqrt {2}} ln {[{ sqrt {2 (x^2 +1)} -x}/{ sqrt {2(x^2 +1)} +x}]}+crond$ .

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