Home/ Intrebari/Q 87892
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

Stefana.
Pe: 2014-10-04T17:57:58+03:00 In: In:

Nevoie de ajutor. Integrale.

1) \int \frac{2tgx}{tg^{2}x+1}dx; x \in \left ( 0, \frac{\pi }{2} \right )

2)\int \frac{\left ( 3-x^{2} \right )dx}{\sqrt{16-x^{2}}}, x\in \left ( -4, 4 \right )

Multumesc anticipat !

Similare

2 raspunsuri

  1. DD
    profesor

    1) Expresia deintegrat ; 2tgx/(1+(tgx)^2)=sin2x.Deci integrala ceruta va fi;
    I=(-cos2x)/2+C
    2)I=integrala[(3-x^2)/√(16-x^2)dx]=3arcsin(x/4)+x.√(16-x^2)-I-13arcsin(x/4)->I=[x.√(16-x^2)]/2-5arcsin(x/4)+C
    I se descompune in doua integrale.Prima se poate face direct, A doua se face prin parti (f=x s g’=x/√(16-x^2)->f’=1 si g=-√(16-x^2))

  2. ghioknt
    profesor

    Avem de calculat \int \frac{2tgx}{tg^{2}x+1}dx; x \in \left ( 0, \frac{\pi }{2} \right ).
    Propun schimbarea de variabila t= tg^2x+1; imi trebuie dt= (tg^2x+1)'dx=2tgx\cdot \frac{1}{\cos^2 x}dx=2tgx( tg^2x+1)dx.
    Deaceea scriu integrala data astfel:
    I=\int \frac{1}{(tg^{2}x+1)^2}\cdot 2tgx( tg^2x+1)dx, iar acum pot sa fac schimbarea anuntata, obtinand:
    J=\int \frac{1}{t^2}dt=-\frac{1}{t}+C.
    Revin la vechea variabila: I=-\frac{1}{tg^2x+1}+C.

Raspunde

Raspunde